2—Infinite Series 28Add these inequalities fromn=kton=∞and you get
f(k) +f(k+ 1) +···>
∫k+1k+
∫k+2k+1+···=
∫∞
kdxf(x)
> f(k+ 1) +f(k+ 2) +···>
∫∞
k+1dxf(x)> f··· (2.9)
The only difference between the infinite series on the left and on the right is one term, so either
everything converges or everything diverges.
You can do better than this and use these inequalities to get a quick estimate of the sum of a
series that would be too tedious to sum by itself. For example
∑∞11
n^2
= 1 +
1
22
+
1
32
+
∑∞
41
n^2
This last sum lies between two integrals.
∫∞3dx
1
x^2
>
∑∞
41
n^2
>
∫∞
4dx
1
x^2
(2.10)
that is, between 1/3 and 1/4. Now I’ll estimate the whole sum by adding the first three terms explicitly
and taking the arithmetic average of these two bounds.
∑∞11
n^2
≈1 +
1
22
+
1
32
+
1
2
(
1
3
+
1
4
)
= 1. 653 (2.11)
The exact sum is more nearly 1.644934066848226, but if you use brute-force addition of the original
series to achieve accuracy equivalent to this 1.653 estimation you will need to take about 120 terms.
This series converges, but not very fast. See also problem2.24.
Quicker Comparison Test
There is another way to handle the comparison test that works very easily and quickly (if it’s applicable).
Look at the terms of the series for largenand see what the approximate behavior of thenthterm is.
That provides a comparison series. This is better shown by an example:
∑∞1n^3 − 2 n+ 1/n
5 n^5 + sinn
For largen, the numerator is essentiallyn^3 and the denominator is essentially 5 n^5 , so for largenthis
series is approximately like
∑∞ 1
5 n^2
More precisely, the ratio of thenthterm of this approximate series to that of the first series goes to
one asn→ ∞. This comparison series converges, so the first one does too. If one of the two series
diverges, then the other does too.
Apply the ratio test to the series forex.
ex=
∑∞
0