1st Year help, inequalities.

akimbo

Hi all, can someone please guide me how to do this question? I have no idea how to approach it :frown:

Thanks

Reply 1

Smaug123

Original post by akimbo

Hi all, can someone please guide me how to do this question? I have no idea how to approach it :frown:

Thanks

Draw a graph, and draw on the appropriate sum in terms of rectangles. What does that look like in comparison with the integral?

Reply 2

akimbo

Original post by Smaug123

Draw a graph, and draw on the appropriate sum in terms of rectangles. What does that look like in comparison with the integral?

Ok thanks. Would this be correct? 1 + 1/4 + 1/9 +...+ 1/n^2 > -1/(n+1) + 1? I'm still kind of confused.

(edited 11 years ago)

Reply 3

DFranklin

Where did you get your expression on the RHS from? (It's not right).

Reply 4

akimbo

Original post by DFranklin

Where did you get your expression on the RHS from? (It's not right).

I integrated 1/x^2 between n+1 and 1 to get the area underneath the curve. Can you explain why it's wrong?

(edited 11 years ago)

Reply 5

DFranklin

Original post by akimbo

I integrated 1/x^2 between n+1 and 1 to get the area underneath the curve. Can you explain why it's wrong?

Wrong sign (which you seem to have found).

General comment: at this level, you should always be "sanity checking" your calculations. It should be obvious that

\int_1^{n+1} \dfrac{1}{x^2}\,dx

is going to be > 0, so your answer can't be negative.

Other than that, it's fine.

Reply 6

akimbo

Original post by DFranklin

Wrong sign (which you seem to have found).

General comment: at this level, you should always be "sanity checking" your calculations. It should be obvious that

\int_1^{n+1} \dfrac{1}{x^2}\,dx

is going to be > 0, so your answer can't be negative.

Other than that, it's fine.

Ok thanks, could you explain how I get from here to show that the statement is true?

Reply 7

akimbo

Am I correct in saying that

\sum_{k=1}^n 1/k^2 = 1

as n

\Rightarrow \infty

Reply 8

Matureb

What are the first few terms?

Reply 9

Smaug123

Original post by akimbo

Am I correct in saying that

\sum_{k=1}^n 1/k^2 = 1

as n

\Rightarrow \infty

No, since even if you take n=2, the sum is already bigger than 1 and it will only increase. This is one of Euler's big results: the value of the sum is in the spoiler.

Spoiler

It's a bit of a pain to prove, although it's easy to show that S, the value of the sum, satisfies S > 1 > S-1, by considering the integral of 1/x^2 from 1 to infinity.
https://en.wikipedia.org/wiki/Basel_problem outlines a proper proof.

Reply 10

akimbo

Original post by Smaug123

No, since even if you take n=2, the sum is already bigger than 1 and it will only increase. This is one of Euler's big results: the value of the sum is in the spoiler.

Spoiler

Oh god I'm so confused. :confused:

Reply 11

davros

Original post by akimbo

Oh god I'm so confused. :confused:

The question's only asking you to put an upper bound on the sum, not evaluate it :smile:

You've virtually done it now: your sum is < 2 - (1/n) for any n > 1, so what will the upper bound be as n-> infinity?

Reply 12

Smaug123

Original post by akimbo

Oh god I'm so confused. :confused:

Ah, I said that "it's easy to show…" and then restated the question you've been given :P
The value of the integral

\int_1^{n}\frac{1}{x^2} dx

is [latex]-\frac{1}{n}+1[\latex], yes? If you draw a graph of y=1/x^2 and draw in the usual bounding rectangles for the integral (above and below the curve) so that the bounds look like the sum you've been given, what can you then say about the sum?

(edited 11 years ago)

Reply 13

akimbo

Original post by davros

The question's only asking you to put an upper bound on the sum, not evaluate it :smile:

You've virtually done it now: your sum is < 2 - (1/n) for any n > 1, so what will the upper bound be as n-> infinity?

Would it be 2? Or was that a stupid answer?

Reply 14

Smaug123

Original post by akimbo

Would it be 2? Or was that a stupid answer?

Quite right, it is 2 :smile:

Reply 15

akimbo

Original post by Smaug123

Quite right, it is 2 :smile:

Ok thanks! Thanks everyone! Just to double check I am right I'll go through my working.

Area underneath the curve between x=1 and x=n+1 is given by

Unparseable latex formula:

\int^{n+1}_1 \frac{1}{x^2}\dx = 1 - \frac{1}{n+1}

[graphs showing upper and lower bounds]

Figure 1 shows that the area of the dotted rectangles is greater than the area under the curve.

Hence,

1 + \frac{1}{4} + \frac{1}{9} +...+ \frac{1}{n^2} > \frac{-1}{n+1} + 1

Which is,

\displaystyle\sum_{k=1}^n \frac{1}{k^2} > 1 - \frac{1}{n+1}

Figure 2 shows the area of the dashed rectangles is less than the area under the curve.

Hence,

\frac{1}{4} + \frac{1}{9} +...+ \frac{1}{n^2} < 1 - \frac{1}{n}

1 + \frac{1}{4} + \frac{1}{9} +...+ \frac{1}{n^2} < 2 - \frac{1}{n}

Therefore,

\displaystyle\sum_{k=1}^n \frac{1}{k^2} < 2 - \frac{1}{n+1}

\displaystyle\sum_{k=1}^n \frac{1}{k^2} < 2 - \frac{1}{n}

for any

n>1

then the upper bound for the infinite series must 2 as n -> infinity.

Would this be correct?

(edited 11 years ago)

Reply 16

Smaug123

Original post by akimbo

Would this be correct?

It is correct, yes - it's probably my fault, but you've done more than is asked for now :smile:

because you've also provided a lower bound of 1, in the "Figure 1" bit. (I know I said to draw in both sets of rectangles, but I suppose it never hurts to have a lower bound too!) You don't actually *need* to have derived

Unparseable latex formula:

\displaystyle\sum_{k=1}^n \frac{1}{k^2} > 1 - \frac{1}{n+1}

for this question.

Reply 17

akimbo

Original post by Smaug123

It is correct, yes - it's probably my fault, but you've done more than is asked for now :smile:

Unparseable latex formula:

\displaystyle\sum_{k=1}^n \frac{1}{k^2} > 1 - \frac{1}{n+1}

for this question.

Thanks a bunch! I'm just glad I've got an answer for this now, couldn't have been possible without the help from everyone in this thread! Many thanks!