How to bound a sum of infinitely many positive real numbers

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How to bound a sum of infinitely many positive real numbers

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[QUICK DESCRIPTION]
This article is about standard tests for convergence of sums of positive real numbers, and how to use them. The most obvious is the comparison test, but others that are sometimes useful are the integral test and the condensation test.

[note article incomplete] More examples needed. So far there are no examples for the integral or condensation tests. It would be nice to do the same series $\sum n^{-2}$ by these two methods. [/note]

[EXAMPLE zeta2]
Here is a well-known proof that the series $\sum_{n=1}^\infty \frac 1{n^2}$ converges. First, the partial sums $S_N=\sum_{n=1}^N\frac 1{n^2}$ form an increasing sequence. Secondly, for each $N\geq 2$, $S_N\leq T_N=1+\sum_{n=2}^N\frac 1{n(n-1)}$ (since $\frac 1{n^2}<\frac 1{n(n-1)}$. And third, $T_N=1+\sum_{n=2}^N(\frac 1{(n-1)}-\frac 1n)=\frac 32-\frac 1N\leq \frac 32$. Therefore, the partial sums $S_N$ form an increasing sequence that is bounded above, which implies that the series converges.

[GENERAL DISCUSSION]
The technique we have just used is called ''the comparison test''. The precise statement of the test is as follows.

[theorem comparison] Let $(a_n)$ and $(b_n)$ be two sequences of non-negative real numbers, and suppose that $a_n\leq b_n$ for every $n$. Then if the series $\sum_{n=1}^\infty b_n$ converges, so does the series $\sum_{n=1}^\infty a_n$. [/theorem]

The proof of the theorem can be abstracted out of the argument in [ref Example #zeta2]. For every $N$, we have $\sum_{n=1}^Na_n\leq\sum_{n=1}^Nb_n\leq\sum_{n=1}^\infty b_n$, so the partial sums of the series $\sum_{n=1}^\infty a_n$ are increasing and bounded above.

It is often useful to have a more flexible version of the comparison test, which follows easily from the version as stated above. Here it is.

[corollary] Let $(a_n)$ and $(b_n)$ be two sequences of non-negative real numbers, and suppose that there exist $M$ and $C>0$ such that $a_n\leq Cb_n$ for every $n\geq M$. Then if the series $\sum_{n=1}^\infty b_n$ converges, so does the series $\sum_{n=1}^\infty a_n$. [/corollary]

To deduce this from [ref Theorem #comparison], observe first that if the series $\sum_{n=1}^\infty b_n$ converges, then so does the series $\sum_{n=1}^\infty Cb_n$. But changing finitely many terms of an infinite series does not affect whether or not it converges, so the series $\sum_{n=1}^{M-1} a_n+\sum_{n=M}^\infty Cb_n$ converges as well. Therefore, by [ref Theorem #comparison], so does the series $\sum_{n=1}^\infty a_n$.

[EXAMPLE]
As an illustration of how to use the slightly more general comparison test, consider the infinite sum $\sum_{n=1}^\infty \frac{29n^2+350n}{n^4-100n^2+1}$. A slightly cheeky, but perfectly rigorous, argument is to say that if $n\geq 1000$, then $29n^2+350n\leq 50n^2$, and $n^4-100n^2+1\geq n^4/2$, so $\frac{29n^2+350n}{n^4-100n^2+1}\leq \frac{100}{n^2}$. Since we have established that $\sum_{n=1}^\infty\frac 1{n^2}$ converges, so does  $\sum_{n=1}^\infty \frac{29n^2+350n}{n^4-100n^2+1}$.

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How to bound a sum of infinitely many positive real numbers

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