Question about limit of series

[JG89]

This isn't a homework question, it's a question I encountered during self-study.

It said prove that the limit as n approaches infinity of 1/n^2 + 1/(n+1)^2 +...+ 1/(2n)^2 = 0

Is it sufficient to say that each term converges to 0 as n approaches infinity, and so the entire sum must also approach infinity?

 

[Ben Niehoff]

No, because infinity*0 is an indeterminate form. You'll have to sum the series and take the limit properly (or perhaps find something more creative).

 

[JeffNYC]

I think you can compare all terms in the series to a P series 1/n^2, which diverges.

Each successive term in the series, a_n < a_n+1 < a_n+2, etc... is declining and all are <= the first term.

So that argument should be OK, without knowing anything else about the series.

Jeff

 

[JG89]

I understand that infinity*0 is indeterminate, but if each term converges to 0, wouldn't the sum look like 0 + 0 + ... + 0? Making the entire thing 0?

 

[Tedjn]

While that appears to be true at first, it is not always true. Consider the function:

f(n) = 1/n + 1/(n+1) + 1/(n+2) + ... + 1/(2n)

The smallest term in this sum is 1/(2n). There are a total of n+1 terms. This means that the sum f(n) > (n+1)/(2n) = 1/2 + 1/(2n) > 1/2 for n >= 1. Thus, it doesn't matter what value of n we have, f(n) is always greater than 1/2.

Yet, in this example, each of the terms tends to 0 as n tends to infinity. What matters is how fast the terms go to 0 compared to how many terms there are.

 

[HallsofIvy]

I think you can compare all terms in the series to a P series 1/n^2, which diverges.

You meant "converges" here, didn't you?

Each successive term in the series, a_n < a_n+1 < a_n+2, etc... is declining and all are <= the first term.

So that argument should be OK, without knowing anything else about the series.

Jeff

 

[Gib Z]

Ben's already implied this, but I though i'd weigh in on this: Try Squeezing it.

 

[JG89]

What matters is how fast the terms go to 0 compared to how many terms there are.

Could you please expand on this?

 

[JG89]

Alright. How about this.

0 < 1/n^2 + 1/(n+1)^2 + ... + 1/(2n)^2 < 1/n^2 + ... + 1/n^2 [There are n+1 terms]

1/n^2 + ... + 1/n^2 = (n+1)/n^2 = 1/n + 1/n + 1/n^2, which converges to 0. Thus the original sequence must also converge to 0.

 

[Gib Z]

Yes, you nailed it.