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### Convergence of Sums

```
Date: 05/07/99 at 12:08:07
From: Shaun Leigh Andrews
Subject: Real Analysis(Convergence of Sums)

I have a problem deducing that the following sum converges absolutely:

Sum from 1 to infinity of (-1)^(n-1)/(n^2)!

I have deduced that it converges since it alternates and
(-1)^(n-1)/(n^2)! -> 0 as n -> infinity.

To test whether the Sum from 1 to infinity of 1/(n^2)! converges (i.e.
test for absolute convergence) I decided to use the ratio test as
below:

Limit as n -> infinity of { (n^2)!/((n+1)^2)!}

However, how do I deduce that this limit is less than 1?

I also wish to estimate the sum to an accuracy of 10^(-5) by computing
upper and lower bounds that differ by less than that amount. How do I
```

```
Date: 05/07/99 at 14:02:35
From: Doctor Wilkinson
Subject: Re: Real Analysis(Convergence of Sums)

Hi, Shaun!

The ratio test is not powerful enough to solve this problem. The limit
in question is actually 1.  You can see this as follows:

n^2/(n+1)^2 = (n/(n+1))^2 = (1/(1+1/n))^2, and
1/n -> 0 as n -> infinity, so the limit is 1/1 or 1.

There are a couple of ways that you can show the sum of 1/n^2
converges, even though the ratio test fails.

The first method is to compare the sum to an integral. This is a very
powerful method and will handle a lot of convergences problems. If you
draw a graph of y = x^2 you can see fairly easily that the sum from
n = 2 to M of 1/n^2 is less than the integral from 1 to M of 1/x^2 dx.
Evaluating this integral will give you an upper bound for the sum.

Another method is to compare the sum to a simliar sum that is easier
to evaluate.  This time we look at

1/2^2 + 1/3^3 + 1/4^2 + ...

and note that it is less than

1/1*2 + 1/2*3 + 1/3*4 +...

but this sum can be written as

1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...

which is what we call a "telescoping" sum:  the terms all cancel
except for the first.

The second part of the problem is to estimate the sum (I assume this
is the original alternating sum) to within 10^(-5). Here you need the
fact that the error in computing an alternating sum is less in
absolute value than the next term after you stop. So you have to sum
just 100 terms (since 100^2 is 10000).

I hope this helps.  If you need more detail, let me know. But you seem
to be fairly on top of things, and I think you can work this out from
here.

Just one minor correction: the theorem about alternating series
requires that the terms be decreasing in absolute value, not just that
they tend to 0.  That's no problem in this example, of course.

- Doctor Wilkinson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Analysis
High School Sequences, Series

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