Analysis on Series

Got two questions I'd like to get checked, thank you.

B2: Since

\displaystyle \sum_{i=1}^{\infty} a_i

diverges to infinity, then the partial sum,

(s_n)

, also diverges to infinity where

\displaystyle (s_n)=\sum_{i=1}^n a_i

By the constant multiple rule for sequences,

c\cdot (s_n)

also diverges for

c>0

therefore

\displaystyle c\cdot (s_n) = c\sum_{i=1}^n a_i = \sum_{i=1}^n c \cdot a_i

diverges therefore

\displaystyle \sum_{i=1}^{\infty} ca_i

diverges.

However, the const multiple rule was covered on convergent sequences and was not proven for divergent ones in my module - it seems intuitively correct but 1) I'm not sure how I would go about proving it if required, and 2) I'm not entirely sure if this is the correct approach to use this.

B4:

(a_n)=\frac{(3x)^n}{n^2}

and the ratio test would be:

\displaystyle \frac{a_{n+1}}{a_n}=\frac{(3x)^{n+1}}{(n+1)^2} \cdot \frac{n^2}{(3x)^n} =\frac{3x\cdot (3x)^n}{n^2+2n+1}\cdot \frac{n^2}{(3x)^n}=\frac{3x\cdot n^2}{n^2+2n+1}=\frac{3x}{1+\frac{2}{n}+\frac{1}{n^2}}

Now as

n\rightarrow \infty

we get

\frac{a_{n+1}}{a_n}=3x

and we want the series to converge therefore this limit must be less than 1.

We get

0\leq 3x < 1 \Rightarrow 0\leq x < \frac{1}{3}

(edited 7 years ago)

Reply 1

joostan

Original post by RDKGames

Got two questions I'd like to get checked, thank you.

B2: Since

\displaystyle \sum_{i=1}^{\infty} a_i

diverges to infinity, then the partial sum,

(s_n)

, also diverges to infinity where

\displaystyle (s_n)=\sum_{i=1}^n a_i

By the constant multiple rule for sequences,

c\cdot (s_n)

also diverges for

c>0

therefore

\displaystyle c\cdot (s_n) = c\sum_{i=1}^n a_i = \sum_{i=1}^n c \cdot a_i

diverges therefore

\displaystyle \sum_{i=1}^{\infty} ca_i

For the first, you can consider a definition of a divergent sequence,

\forall K \in \mathbb{R}^{+}, \ \exists N(K) \in \mathbb{N}

such that

\forall n \geq N, \ \ \ |s_n| > K

.
From here it's easy enough to get your result.

The second looks OK, though I've not checked it.
You've actually missed the case with

x=\frac{1}{3}

which is also convergent - this is the Basel problem.

(edited 7 years ago)

Reply 2

atsruser

Original post by joostan

The second looks OK, though I've not checked it.
You've actually missed the case with

x=\frac{1}{3}

which is also convergent - this is the Basel problem.

That doesn't follow from the ratio test though.

Reply 3

joostan

Original post by atsruser

That doesn't follow from the ratio test though.

True, but the question asks you to find values for which it converges. . .
Besides, if memory serves, isn't the ratio test is inconclusive if the ratio is 1, and in the interest of being thorough, one ought to check this case separately?

(edited 7 years ago)

Reply 4

RDKGames

Study Forum Helper

Original post by joostan

For the first, you can consider a definition of a divergent sequence,

\forall K \in \mathbb{R}^{+}, \ \exists N(K) \in \mathbb{N}

such that

\forall n \geq N, \ \ \ |s_n| > K

.
From here it's easy enough to get your result.

The second looks OK, though I've not checked it.
You've actually missed the case with

x=\frac{1}{3}

which is also convergent - this is the Basel problem.

Ah thank you - I got the proof for the const. mult. rule from that as required.

I'm unsure about

x=\frac{1}{3}

though as the test is inconclusive but I can see where you're coming from, and it works out by plugging it back through.

Reply 5

joostan

Original post by RDKGames

Ah thank you - I got the proof for the const. mult. rule from that as required.

I'm unsure about

x=\frac{1}{3}

though as the test is inconclusive but I can see where you're coming from, and it works out by plugging it back through.

No problem :smile:

Yeah. Personally, I would include the additional case, especially as it is a well known result and isn't tricky to prove; but I guess there is some ambiguity in the phrasing of the question.

Reply 6

Zacken

It's not hard to do the Basel problem convergence, simply compare it to

\displaystyle 1 + \sum_{n>1} \frac{1}{n(n-1)}

which telescopes.

Reply 7

atsruser

Original post by joostan

True, but the question asks you to find values for which it converges. . .

Well, the question is: "Use the ratio test to establish for which values of x >=0 the series <blah > converges".

So it implies to my mind that only the results from the ratio test are required. However, the wording is a bit off, I'd say, since the ratio test can't in fact establish what is being asked. I'd prefer it to be something like: "For which values of x>=0 does the ratio test allow us to conclude that the series <blah> converges".

Besides, if memory serves, isn't the ratio test is inconclusive if the ratio is 1, and in the interest of being thorough, one ought to check this case separately?