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As an example application of the comparison test, Spivak introduces the series $\displaystyle \sum_{n=1}^{\infty} \frac{n+1}{n^{2}+1}$. He says, "we would expect this series to diverge, since $\frac{n+1}{n^{2}+1}$ is practically $\frac{1}{n}$ for large $n$. To prove this, choose any number $c$ with $0<c<1$. Then

$$0 \leq \frac{1}{n} < c \frac{n+1}{n^{2}+1}$$

for large enough $n$."

Here's my attempt at reproducing this inequality.

Since $\frac{1}{n} <\frac{n+1}{n^{2}+1}$, we have $\frac{\frac{1}{n}}{\frac{n+1}{n^{2}+1}} <1$.

Combined with the fact that $\displaystyle \lim_{n=1}^{\infty} \frac{\frac{1}{n}}{\frac{n+1}{n^{2}+1}} =1$, we see that the quotient of the two terms increases to 1 as $n$ increases.

In particular, for $0<c<1$,

$$\frac{\frac{1}{n}}{\frac{n+1}{n^{2}+1}} >c \text{ for large enough }n.$$

So $\frac{1}{n} > c \frac{n+1}{n^{2}+1}$.

However, this is the opposite of the inequality mentioned in the text. Could you help me to see what I'm missing?

joeshiki
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2 Answers2

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To be honest I have no idea why we should choose $0<c<1.$ One should use a large constant, say $c= 100$. Then it is easy to verify that $$1/n < 100 \frac{n+1}{n^2+1}.$$ Indeed, let $f(n) = n^2+1 - 100n(n+1).$ Note that to prove the above inequality, it is enough to show $f(n)<0$ (at least for all large $n$). Since $f(n) = -99n^2 - 100n + 1$, the assertion is trivial; we have $f(n)<0$ for $n\ge1.$

You can then apply the comparison test, since $$\sum \frac{1}{100n}$$ is divergent.

Ssay
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Note that $\lim_{n \to \infty} {{1 \over n} \over {n+1 \over n^2+1}} =1$, go for any $c>1$ there is some $N$ such that ${1 \over n} \le c {n+1 \over n^2+1}$ for all $n \ge N$.

copper.hat
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