Try splitting the right hand side into partial fractions and finding the partial sum up to N. Then let N go to infinity and see what happens to the infinite sum.

Next use comparison test :smile:

Let me know how it goes.

\frac{1}{n}

is harder, the easiest method being the integral test.

The integral test is usually a good shout for problems of this sort. Obviously you can prove them without the integral test, usually by some clever comparison, but they're often hard to find.

(edited 9 years ago)

Reply 3

username1050473

14

Original post by TheAlfie3000

Is it not just as simple as |r|<1?

Posted from TSR Mobile

If you consider the large values of n, you see that you get smaller and smaller numbers:

n=1, s = 1.
n = 10, s = 1/10.

Reply 4

username1192424

OP

4

Can I link it to the integrals the fact that the integral tends to a value as x reaches infinity? Or is this completely nonsense?

Reply 5

Smaug123

15

Original post by sangyoonknow

Can I link it to the integrals the fact that the integral tends to a value as x reaches infinity? Or is this completely nonsense?

This is correct, and is known as the integral test. The standard way is known as the Cauchy condensation test: group together consecutive

2^m

terms, to get

1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots \geq (\frac{1}{2})+(\frac{1}{2})+( \frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+ \frac{1}{8}+ \frac{1}{8})+\dots

, which clearly diverges, being just infinitely many

\frac{1}{2}

s. (The grouping is valid, since the summand is always positive, so if the series converges then it converges absolutely.)

Original post by sarcasmrules

If you consider the large values of n, you see that you get smaller and smaller numbers:

n=1, s = 1.
n = 10, s = 1/10.

This is a necessary, but insufficient, condition for convergence.

\sum \frac{1}{n}

diverges.

Reply 6

poorform

10

A series of the form

\displaystyle \sum \frac{1}{n^{\alpha }}

converges for

\displaystyle \alpha > 1

and diverges elsewhere. To go back to your original question which is does

\displaystyle \sum \frac{1}{n^{2}}

converge or diverge, well it converges by the rule above. You can prove this by proving that the partial sum is bounded above and since

\displaystyle \sum \frac{1}{n^{2}}

is a sequence of non-negative terms that it is also increasing and therefore convergent by the monotone convergence theorem.

Reply 7

Smaug123

15

Original post by poorform

A series of the form

\displaystyle \sum \frac{1}{n^{\alpha }}

converges for

\displaystyle \alpha > 1

and diverges elsewhere. To go back to your original question which is does

\displaystyle \sum \frac{1}{n^{2}}

converge or diverge, well it converges by the rule above. You can prove this by proving that the partial sum is bounded above and since

\displaystyle \sum \frac{1}{n^{2}}

is a sequence of non-negative terms that it is also increasing and therefore convergent by the monotone convergence theorem.

I'd clarify this by saying that the partial sum is bounded above *by an appropriate integral*; otherwise I don't see an easy way to notice that the partial sums are bounded above. (I'd be interested to know of one - it's quite possible that I've just forgotten it.)

ETA: my bad, the partial-fractions way outlined by TheIrrational does this fine.

(edited 9 years ago)

Reply 8

DFranklin

18

Original post by Smaug123

I'd clarify this by saying that the partial sum is bounded above *by an appropriate integral*; otherwise I don't see an easy way to notice that the partial sums are bounded above. (I'd be interested to know of one - it's quite possible that I've just forgotten it.)

ETA: my bad, the partial-fractions way outlined by TheIrrational does this fine.

TheIrrational

..

Except for the trivial issue that it's not true that

\dfrac{1}{n^2} < \dfrac{1}{n(n+1)}

.

[The irony being that it actually is a fairly trivial issue to fix, it's just requires an annoying bit of fiddling.

My preference was to go: "if n > 1,

\dfrac{1}{n^2} < \dfrac{1}{n(n-1)} = \dfrac{1}{n-1} - \dfrac{1}{n}

" - then the only bit of fiddling is the "n>1" at the start (to avoid the division by 0)].

Reply 9

TheIrrational

15

Original post by DFranklin

Except for the trivial issue that it's not true that

\dfrac{1}{n^2} < \dfrac{1}{n(n+1)}

.

[The irony being that it actually is a fairly trivial issue to fix, it's just requires an annoying bit of fiddling.

My preference was to go: "if n > 1,

\dfrac{1}{n^2} < \dfrac{1}{n(n-1)} = \dfrac{1}{n-1} - \dfrac{1}{n}

" - then the only bit of fiddling is the "n>1" at the start (to avoid the division by 0)].

Ah yes, I did mean to write \frac{1}{n(n-1)}. Thank you for pointing this out and explaining that the n>1 needs to be considered. I forgot this! :smile:

How to solve whether infinite sum of 1/x^2 converges or diverges?

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