STEP 2003 Solutions Thread Watch

SimonM
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STEP III, Question 2

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\displaystyle \binom{2r}{r}  = \frac{2r!}{r!r!} = \frac{1 \times 2 \times 3 \times 4 \times \cdots (2r-1) \times 2r}{r! r!} = \frac{1 \times 3 \times 5 \cdots (2r-1) \times 2 \times 4 \cdots 2r}{r!r!} = \frac{1 \times 3 \times 5 \cdots (2r-1) \times 2^r r!}{r!r!} = \frac{1 \times 3 \times 5 \cdots (2r-1)}{r!} \times 2^r

\displaystyle (1-p)^{-1/2} =

\displaystyle 1 - \frac{1}{2} (-p) \, + \, \left ( - \frac{1}{2} \right) \left ( - \frac{3}{2}\right) \frac{p^2}{2!} \, +
\displaystyle \left ( - \frac{1}{2} \right) \left ( - \frac{3}{2}\right) \left ( -\frac{5}{2} \right ) \frac{p^3}{3!} \, + \, \left ( - \frac{1}{2} \right) \left ( - \frac{3}{2}\right) \left ( -\frac{5}{2} \right ) \left ( -\frac{7}{2} \right )\frac{p^4}{4!} \, + \, \ldots =

\displaystyle 1\, + \,\frac{1}{1!} \left ( \frac{p}{2} \right) \, + \, \frac{1 \times 3}{2!} \left ( \frac{p}{2} \right )^2 \, + \, \frac{1 \times 3 \times 5}{3!} \left ( \frac{p}{2} \right)^3 \, + \, \frac{1 \times 3 \times 5 \times 7}{4!} \left ( \frac{p}{2} \right)^4 \, + \, \ldots =

\displaystyle \binom{0}{0} \frac{1}{2^0} \left ( \frac{p}{2} \right )^0 +  \binom{2}{1} \frac{1}{2^1}  \left ( \frac{p}{2} \right)  +  \binom{4}{2} \frac{1}{2^2} \left ( \frac{p}{2} \right )^2  + \binom{6}{3} \frac{1}{2^3} \left ( \frac{p}{2} \right )^3  + \binom{8}{4} \frac{1}{2^4} \left ( \frac{p}{2} \right )^4 + \ldots =

Set p = 1/2 to get the identity in the question

\displaystyle \sum_{r=0}^{\infty} \binom{2r}{r} x^r  = (1-4x)^{-1/2}

Differentiating with respect to x and multiplying by 2x and adding the original expression

\displaystyle \sum_{r=0}^{\infty} (2r+1) \binom{2r}{r} x^r = \frac{1}{(1-4x)^{3/2}}

Putting x = 1/5 we get the required identity
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Unbounded
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(Original post by tommm)
n.b. I don't know how to LaTex the dots denoting differentiation w.r.t to time, so I'm just using x for displacement, v for speed and a for acceleration.
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Do you mean like this:  \dot{x} and  \ddot{x} ?

The codes are \dot{x} and \ddot{x} respectively.
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SimonM
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STEP III, Question 6

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A trivial consequence of the product to sum formulae

Let S be the sum

\displaystyle 2 \sin \frac{1}{2} \theta S = \left ( \sin \left( a + \frac{1}{2}\right ) \theta - \sin \left ( a - \frac{1}{2}\right ) \right ) + \left ( \sin \left( a +1+ \frac{1}{2}\right ) \theta - \sin \left ( a 

+1- \frac{1}{2}\right ) \right ) + \cdots + \left ( \sin \left( b-1+ \frac{1}{2}\right ) \theta - \sin \left ( b - 1-\frac{1}{2}\right ) \right ) =\sin \left( b- \frac{1}{2}\right ) - \sin \left( a- \frac{1}{2}\right )

So

\displaystyle S = \dfrac{\sin \left( b- \frac{1}{2}\right )\theta - \sin \left( a- \frac{1}{2}\right )\theta}{2 \sin \frac{1}{2} \theta} = \dfrac{\sin \left ( \frac{(b-a)\theta}{2}\right ) \cos \left ( \frac{(b+a-1)\theta}{2} \right )}{\sin \frac{1}{2} \theta} = 0

Therefore (b-a)\theta = 2n\pi or (b+a-1)\theta = (2n+1) \pi
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DFranklin
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(Original post by Horizontal 8)
STEP II Q3

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As  n \to \infty
 \frac{1}{3^{n}} \to 0
 \implies U_n \to 1 As n increases without bound.

(Therefore the given integer m is one.)

I grant you it's not 100% clear, but I would interpret this part as "given an integer m, give an example of a sequence of irrational numbers that converges to m". (In other words, you have to produce a sequence that works for every m).
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Horizontal 8
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(Original post by DFranklin)
I grant you it's not 100% clear, but I would interpret this part as "given an integer m, give an example of a sequence of irrational numbers that converges to m". (In other words, you have to produce a sequence that works for every m).
Define a sequence  A_n = m+U_n-1 of which all the terms are clearly irrational since U_n is irrational for all n and a rational number + irrational number is always irrational. Yet it converges to m as  n \to \infty

Is this what you mean?
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qgujxj39
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STEP II 2003 Q1

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i)

when a = 0,

-y - z = 3 (1)
-y - 3z = 7 (2)
-y - 5z = b (3)

(1) - (2) gives 2z = -4 \implies z = -2
and also y = -1

Therefore, a solution exists iff -(-1) -5(-2) = b \implies b = 11

ii)

ax -y - \lambda = 3 (1)
2ax -y - 3\lambda = 7 (2)
3ax -y - 5\lambda = b (3)

(2) - 2.(1) gives
y - \lambda = 1 \implies y = 1 + \lambda
Substituting this into (1) gives x = \frac{4 + 2\lambda}{a}

Substituting this values into (3) shows that the solution is consistent no matter what \lambda equals.

iii)
When a = 2, x = 2 + \lambda, y = 1 + \lambda, z = \lambda

therefore x^2 + y^2 + z^2 = (2 + \lambda)^2 + (1 + \lambda)^2 + \lambda^2 = 3\lambda^2 + 6\lambda + 5 = 3(\lambda + 1)^2 + 2

This is least when the squared part equals 0, i.e. \lambda = -1

Therefore x = 1, y = 0, z = -1

iv)

If \lambda = -0.5, then y^2 + z^2 = 2 \times 0.25 < 1

and x = \frac{3}{a}

Therefore we can take something like a = 10^{-10} to give x > 10^6, and this has a solution which satisfies both conditions.


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qgujxj39
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(Original post by GHOSH-5)
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Do you mean like this:  \dot{x} and  \ddot{x} ?

The codes are \dot{x} and \ddot{x} respectively.
Thank you, I shall remember that
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Horizontal 8
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(Original post by tommm)
STEP II 2003 Q1

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i)

when a = 0,

-y - z = 3 (1)
-y - 3z = 7 (2)
-y - 5z = b (3)

(1) - (2) gives 2z = -4 \rightarrow z = -2
and also y = -1

Therefore, a solution exists iff -(-1) -5(-2) = b \rightarrow b = 11

ii)

ax -y - \lambda = 3 (1)
2ax -y - 3\lambda = 7 (2)
3ax -y - 5\lambda = b (3)

(2) - 2.(1) gives
y - \lambda = 1 \rightarrow y = 1 + \lambda
Substituting this into (1) gives x = \frac{4 + 2\lambda}{a}

Substituting this values into (3) shows that the solution is consistent no matter what \lambda equals.

iii)
When a = 2, x = 2 + \lambda, y = 1 + \lambda, z = \lambda

therefore x^2 + y^2 + z^2 = (2 + \lambda)^2 + (1 + \lambda)^2 + \lambda^2 = 3\lambda^2 + 6\lambda + 5 = 3(\lambda + 1)^2 + 2

This is least when the squared part equals 0, i.e. \lambda = -1

Therefore x = 1, y = 0, z = -1

iv)

If \lambda = -0.5, then y^2 + z^2 = 2 \times 0.25 < 1

and x = \frac{3}{a}

Therefore we can take something like a = 10^{-10} to give x > 10^6, and this has a solution which satisfies both conditions.



For implications I'd use \implies and \iff since your \rightarrow is coming out as a 'tends to' (or \to) on my screen atleast.
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SimonM
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You could us \Rightarrow (note the capital)
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DFranklin
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(Original post by Horizontal 8)
Define a sequence  A_n = m+U_n-1 of which all the terms are clearly irrational since U_n is irrational for all n and a rational number + irrational number is always irrational. Yet it converges to m as  n \to \infty

Is this what you mean?
Yes. I doubt it would lose you much here, because it's so trivial to fix it to work for the general case. But the standard interpretation of "show XXX for a given integer n" means "it has to work for any n, but you can consider n as fixed".
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qgujxj39
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(Original post by Horizontal 8)
For implications I'd use \implies and \iff since your \rightarrow is coming out as a 'tends to' (or \to) on my screen atleast.
Cheers, I'll clean that up.
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Horizontal 8
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(Original post by DFranklin)
Yes. I doubt it would lose you much here, because it's so trivial to fix it to work for the general case. But the standard interpretation of "show XXX for a given integer n" means "it has to work for any n, but you can consider n as fixed".
ok thanks
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qgujxj39
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STEP II 2003 Q2

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\theta = \pi/3

Let 2\sqrt{3}\sin\theta + 4\cos\theta = R\cos(\theta + \alpha)

We find, using the usual method, that R = \sqrt{28}, \alpha = -\arctan(\frac{\sqrt{3}}{2})

Therefore \sqrt{28}\cos(\theta - \arctan(\frac{\sqrt{3}}{2})) = 5
\implies \theta = \arccos\frac{5}{\sqrt{28}} + \arctan\frac{\sqrt{3}}{2}

Now remembering that \theta = \pi/3, and multiplying both sides by 3, we achieve the required result.

For the second part, by comparison with the first, we want to find an equation of a similar form that leads to
10\sin(\theta + \arctan\frac{3}{4}) = 7\sqrt{2}

Using a bit of guesswork, we find that this equation is
8\sin\theta + 6\cos\theta = 7\sqrt{2}
which is satisfied by \theta = \pi/4

Rearranging 10\sin(\theta + \arctan\frac{3}{4}) = 7\sqrt{2}, we get:

\theta = \arcsin\frac{7\sqrt{2}}{10} - \arctan\frac{3}{4}

and remembering that \theta = \pi/4 leads to the required result.


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Horizontal 8
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STEP I question 2

First part:
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 \displaystyle \frac{1}{a+b} = \frac{1}{a} + \frac{1}{b} \iff  \frac{1}{a+b} -\frac{1}{a} - \frac{1}{b} =0

 \displaystyle \iff \frac{ab-b(a+b)-a(a+b)}{ab(a+b)} = 0

 \displaystyle \iff \frac{-b^2-a^2-ab}{ab(a+b)} = 0

 \displaystyle \iff a^2+b^2+ab =0

Considering the discriminant:
In order for a to be real  b^2-4b^2 =-3b^2\geq 0
Which is impossible for real values of b
Similarly: In order for a to be real  a^2-4a^2 =-3a^2\geq 0
Which is impossible for real values of a

Therefore there are no values of a and b for which this will give the correct answer.



Second part:
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 \displaystyle \frac{1}{a+b+c} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \iff  \frac{1}{a+b+c} -\frac{1}{a} - \frac{1}{b} -\frac{1}{c}=0

Putting it all under a common denominator:
 \displaystyle \iff \frac{abc-bc(a+b+c)-ac(a+b+c)-ab(a+b+c)}{abc(a+b+c)} =0

Multiplying out:
 \iff abc-abc-b^2c-bc^2-a^2c-abc-ac^2-a^2b-ab^2-abc = 0

 \iff 2abc-b^2c-bc^2-a^2c-ac^2-a^2b-ab^2=0

 \iff (a+b)(b+c)(c+a) = 0

Hence the answer will be correct if and only if one of the three equations hold.
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Unbounded
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Question 11, STEP I, 2003
First Part
Let the particle starting at A be labelled P and the particle starting at B be labelled Q. Their masses respectively are  m_P and  m_Q . Let A be a distance l away from where P and Q meet. Let the plane be inclined at angle  \theta to the horizontal. Let T be the time when they reach after setting off from A and B.

We can find their accelerations:

 m_Pg\sin \theta = m_P a_P

 \implies a_P = g\sin \theta

Similarly Q accelerates  g\sin \theta down the slope.

Using SUVAT parallel to the plane:

For P:
 l = uT - \frac{1}{2}g\sin \theta T^2

For Q:
 d-l = \frac{1}{2}g\sin \theta T^2

Adding these equations gives:

 d = uT \implies \boxed{T = \frac{d}{u}} \ \ \ \square
Second Part
We can find their speeds just before the collision using SUVAT:

For P:
 v = u - \frac{dg\sin \theta}{u} = \frac{u^2-dg\sin \theta}{u}

For Q:
 v = \frac{dg\sin \theta}{u}

We can now find the kinetic energy before the collision:
 KE_{\mathrm{before}} = \frac{1}{2} m \left( \frac{u^2-dg\sin \theta}{u} \right)^2 + \frac{1}{2}m\left(\frac{dg\sin \theta}{u}\right)^2

 KE_{\mathrm{before}} = \frac{1}{2}m \left(\dfrac{u^4-2u^2dg\sin \theta + 2d^2g^2\sin^2 \theta}{u^2} \right)

We need to find the speeds of P and Q after the collision. Let their speeds after the collision be  v_P and  v_Q respectively. Let us assume their directions of motions change after the collision.
By Newton's law of restitution, we find that:
 eu = v_P + v_Q

And by the conservation of linear momentum, we find that:

 \dfrac{u^2-2dg\sin \theta}{u} = -v_P + v_Q

Subtracting these two equations, we find that:

 v_P = \dfrac{u^2(e-1) + 2dg\sin \theta}{2u}

 \implies v_P^2 = \dfrac{u^4(e-1)^2 + 4u^2(e-1)dg\sin \theta + 4d^2g^2\sin^2 \theta}{4u^2}

And adding those two equations, we find that:

 v_Q = \dfrac{u^2(e+1) -2dg\sin \theta}{2u}

 \implies v_Q^2 = \dfrac{u^4(e+1)^2 -4u^2(e+1)dg\sin \theta + 4d^2g^2 \sin^2 \theta}{4u^2}

We can now find the KE after the collision:

 KE_{\mathrm{after}} = \frac{1}{2}m(v_P^2 + v_Q^2)

 KE_{\mathrm{after}} = \frac{1}{2}m \left(\dfrac{2u^4(e^2+1)+8u^2dg\  sin \theta + 8d^2g^2\sin^2 \theta}{4u^2}\right)

And we can now find the loss in KE due to the collision:
 KE_{\mathrm{loss}} = KE_{\mathrm{before}} - KE_{\mathrm{after}}

 = \frac{1}{2}m \left(\dfrac{u^4-2u^2dg\sin \theta + 2d^2g^2\sin^2 \theta}{u^2} \right) - \frac{1}{2}m \left(\dfrac{2u^4(e^2+1)+8u^2dg\  sin \theta + 8d^2g^2\sin^2 \theta}{4u^2}\right)

 = \frac{1}{2}m \left(\dfrac{u^4(4-2(e^2+1))}{4u^2}\right)

 = \frac{1}{2}mu^2 \left(\dfrac{2-2e^2}{4}\right)

 = \boxed{\frac{1}{4}mu^2(1-e^2)} \ \ \ \square
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SimonM
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STEP II, Question 7

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\displaystyle \int_{e^{1/n}}^{\infty} \frac{\ln x}{x^{n+1}} \, dx = \left [ -\frac{\ln x}{nx^n} \right ]_{e^{1/n}}^{\infty} + \int_{e^{1/n}}^{\infty} \frac{1}{nx^{n+1}} \, dx = \frac{2}{en^2}

If 1<a<b then

\displaystyle \int_a^b \frac{\ln x}{x^{n+1}} \, dx > 0 since we are integrating a strictly positive, function, the conclusion follows.

\displaystyle \sum_{n=1}^N \frac{1}{n^2} = \frac{e}{2}\sum_{n=1}^N \frac{2}{en^2} = \frac{e}{2} \sum_{n=1}^{N} \int_{e^{1/n}}^{\infty} \frac{\ln x}{x^{n+1}} \, dx < \frac{e}{2} \int_{e^{1/N}}^{\infty} \sum_{n=1}^{N} \frac{\ln x}{x^{n+1}} \, dx = \frac{e}{2} \int_{e^{1/N}}^{\infty} \frac{1-x^{-N}}{x^2-x} \ln x \, dx
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qgujxj39
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STEP II 2003 Q4

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The required area is given by

2\displaystyle\int^R_d\sqrt{R^2-y^2}\mathrm{d}y, which can be seen from drawing a diagram[/latex]

Upon using the substitution y = R\cos u, and some manipulation with the cosine double-angle formula, we get

R^2(u - 0.5\sin(2u)|^{\arccos(d/R)}_0 (N.B. that represents limits of integration)

The difficulty in evaluating this lies in the evaluation of 0.5\sin(2\arccos(d/R)). First, using the sine double angle formula, we get

0.5\sin(2\arccos(d/R)) = \sin(\arccos(d/R))\cos(\arccos(d/R)) = (d/R)\sin(\arccos(d/R))

By drawing a suitable right-angled triangle, we find that \sin(2 \arccos (d/R)) = \frac{\sqrt{R^2 - d^2}{R}} and from this we can find the required result.


To find the area in the three listed cases, we note that i) and ii) are both specific versions of iii), therefore we need only find the answer to iii) and i) and ii) follow easily.

From a diagram, we see that we must replace d with the perpendicular distance from the line to the furthest point on the circle from the line. This gives us:

d + D = R, where D is the shortest distance from the origin to the line. The distance from a point (x, y) to the centre is given by

D^2 = (x - a)^2 + (y - b)^2

But the point (x, y) lies on the line y = mx + c, therefore

D^2 = (x - a)^2 + (mx + c - b)^2

To find the minimum value of D, we must differentiate this expression and set it equal to 0. This gives us

x - a + m^2x + mc - mb = 0

\implies x = \displaystyle\frac{mb - mc + a}{m^2 + 1}

substituting this back into our expression for D^2, we get

D^2 = \displaystyle\frac{(mb - mc - m^2a)^2 + (ma + c - b)^2}{(m^2 + 1)^2}

and therefore, to find the new area, we must modify (*) such that

d = R - \sqrt{\displaystyle\frac{(mb - mc - m^2a)^2 + (ma + c - b)^2}{(m^2 + 1)^2}}

The answer to i) follows by letting m = 0; the answer to ii) follows by letting a = b = 0.

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Unbounded
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Question 6, STEP I, 2003
(i)
Case 1: a = 2

 \displaystyle\int_0^1 \dfrac{1}{(x+2)^2} \ \mathrm{d}x = \left [ \dfrac{-1}{x+2} \right ] _0^1

 = -\dfrac{1}{3} + \dfrac{1}{2} = \boxed{\dfrac{1}{6}}

Case 2: a ≠ 2

 \displaystyle\int_0^1 \dfrac{1}{(x+2)(x+a)} \ \mathrm{d}x = \dfrac{1}{a-2} \displaystyle\int_0^1 \left( \dfrac{1}{x+2} - \dfrac{1}{x+a} \right) \ \mathrm{d}x

 = \dfrac{1}{a-2} \left [ \ln (x+2) - \ln (x+a) \right ] _0^1

 = \dfrac{1}{a-2} \left [ \ln 3 - \ln (a+1) -\ln 2 + \ln a \right ]

 = \boxed{\dfrac{1}{a-2} \ln \left( \dfrac{3a}{2(a+1)} \right)}
(ii)
With the substitution  u = x+1 we find this second integral is exactly the same as the first, and so we have the same two cases with the same two answers.
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qgujxj39
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(Original post by DFranklin)
I grant you it's not 100% clear, but I would interpret this part as "given an integer m, give an example of a sequence of irrational numbers that converges to m". (In other words, you have to produce a sequence that works for every m).
Sorry to pester you, but when it says "converges to a given integer m", does that mean that its sum converges to m or does that mean the general term tends to m?
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SimonM
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(Original post by tommm)
Sorry to pester you, but when it says "converges to a given integer m", does that mean that its sum converges to m or does that mean the general term tends to m?
If a sequence converges to a value, then the general term converges to that value
If a series or sum converges to a value then the sum of the series converges to that value
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