Compare Infinite Sums 2

Calculus Level 3

Which is larger?

A. $\begin{aligned} \sum_{\substack{n>1\\m>1}}^{\infty} \frac{1}{n^{m}} \end{aligned}$

B. $\begin{aligned} \sum_{\substack{n>1\\m>1}}^{\infty} \frac{1}{n^{m}-1} \end{aligned}$

Note:

In the above infinite series, $n$ and $m$ are positive integers greater than 1.
Double counting IS allowed in the first infinite series, i.e. if $n^{m}=m^{n}$ , then both $\frac{1}{n^{m}}$ and $\frac{1}{m^{n}}$ appear in the series. For example, since $2^{4}=4^{2}$ , you would count $\frac{1}{2^{4}}=\frac{1}{4^{2}}=\frac{1}{16}$ twice.
Double counting IS NOT allowed in the second infinite series, i.e. if $n^{m}=m^{n}$ , then only one of $\frac{1}{n^{m}-1}$ and $\frac{1}{m^{n}-1}$ appear in the series. For example, since $2^{4}=4^{2}$ , you would only count $\frac{1}{2^{4}-1}=\frac{1}{4^{2}-1}=\frac{1}{15}$ once .

Equal B A

1 solution

Joshua Lowrance
Oct 8, 2018

(I use the fact that $\sum_{n=1}^{\infty} a^{n} = \frac{1}{a-1}$ )

$\begin{aligned} \sum_{\substack{n>1\\ m>1}}^{\infty} \frac{1}{n^{m}} &=(\frac{1}{2^{2}} + \frac{1}{2^{3}} + \frac{1}{2^{4}} + \cdots) + (\frac{1}{3^{2}} + \frac{1}{3^{3}} + \frac{1}{3^{4}} + \cdots) + (\frac{1}{4^{2}} + \frac{1}{4^{3}} + \frac{1}{4^{4}} + \cdots) +\cdots \\ &=(\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \frac{1}{2^{4}} + \cdots) - \frac{1}{2} + (\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \frac{1}{3^{4}} + \cdots) - \frac{1}{3} + (\frac{1}{4} + \frac{1}{4^{2}} + \frac{1}{4^{3}} + \frac{1}{4^{4}} + \cdots) - \frac{1}{4}+\cdots \\\\ &=1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots \\\\\\ &=1 \\\\\\\\ \end{aligned}$

$\begin{aligned} \sum_{\substack{n>1\\m>1}}^{\infty} \frac{1}{n^{m}-1} &=\frac{1}{2^{2}-1}+\frac{1}{2^{3}-1}+\frac{1}{3^{2}-1}+\frac{1}{2^{4}-1}+\cdots \\ &=\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{15}+\cdots \\\\ \end{aligned}$

On a slightly different note,

$\begin{aligned} S_{1} &=\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots \\ &=\frac{1}{1} + (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots) + (\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots) + (\frac{1}{5} + \frac{1}{25} + \frac{1}{125} + \cdots) + \cdots \\\\ \end{aligned}$

We would leave out every $\frac{1}{n^{m}} + \frac{1}{n^{m^{2}}} + \frac{1}{n^{m^{3}}} + \cdots$ because we already covered all those terms in $\frac{1}{n} + \frac{1}{n^{2}} + \frac{1}{n^{3}} + \cdots$ . For instance, we would leave out $\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \cdots$ because $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$ already covers all those terms, and we don't want to repeat terms. This way, we cover every term once, and only once.

$\begin{aligned} S_{1} &=\frac{1}{1} + (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots) + (\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots) + (\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \cdots) + (\frac{1}{5} + \frac{1}{25} + \frac{1}{125} + \cdots) + \cdots - (\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \cdots) - (\frac{1}{8} + \frac{1}{64} + \frac{1}{256} + \cdots) - (\frac{1}{9} + \frac{1}{81} + \frac{1}{729} + \cdots) - (\frac{1}{16} + \frac{1}{256} + \frac{1}{4096} + \cdots) - \cdots \\ &=\frac{1}{1} + \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots - \frac{1}{3} - \frac{1}{7} - \frac{1}{8} - \frac{1}{15} - \cdots \\\\ &=\frac{1}{1} + S_{1} - \frac{1}{3} - \frac{1}{7} - \frac{1}{8} - \frac{1}{15} - \cdots \\\\\\ \end{aligned}$

$0 = 1 - \frac{1}{3} - \frac{1}{7} - \frac{1}{8} - \frac{1}{15} - \cdots$

$\frac{1}{3} + \frac{1}{7} + \frac{1}{8} + \frac{1}{15} + \cdots = 1$

Therefore, A = B = 1.

But in the proof of your second infinite sum, you have started of with S-1 which is actually undefined....!!!!

Aaghaz Mahajan - 2 years, 8 months ago

Compare Infinite Sums 2

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