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Given the following infinite series:

$\sum_{n=1}^{\infty}\frac{1}{(n)^{p}}-\frac{1}{(n+1)^{p}}$

Find the summation of this series?

My approach:

The above series expands to:

Sum = $(\frac{1}{1^p}-\frac{1}{2^p}) +(\frac{1}{2^p}-\frac{1}{3^p}) + (\frac{1}{3^p}-\frac{1}{4^p})+ (\frac{1}{4^p}-\frac{1}{5^p}).... (\frac{1}{(\infty)^p}-\frac{1}{(\infty+2)^p})$

Since, from second term onward, all neighboring terms cancel each other out all the 'middle' terms will cancel each other out. Thus it will evaluate to:

Sum = $(\frac{1}{1^p} - \frac{1}{(\infty+2)^p})$

As, for any finite real/complex value of: $p$ $(\infty+2)^p = \infty$. Thus,

Sum = $(\frac{1}{1^p} - \frac{1}{\infty})$.

As, $\frac{1}{\infty} = 0$. Thus:

Sum = $\frac{1}{1^p}$. (for any finite value of $p$).

Query: Is the above argument correct and can we can replace the series with $\frac{1}{1^p}$ in any equation or not?

TheoryQuest1
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1 Answers1

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Yes you can, although the symbol $\infty$ can be handled in a different way. For example, call

$$ S_N = \sum_{n=1}^N\frac{1}{n^p} - \frac{1}{(n + 1)^p} $$

You showed that

$$ S_N = 1 - \frac{1}{(N + 1)^p} $$

The sum you are interested in is

$$ S = \lim_{N \to \infty} S_N = 1 $$

caverac
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  • Thank you. As I understand, when we say $S = \lim_{N \to \infty} S_N = 1$ instead of $\frac{1}{1^p}$ we essentially imply that as $\lim_{N \to \infty} then \frac{1}{(N + 1)^p} \to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case? – TheoryQuest1 Dec 19 '18 at 12:52
  • (2) In any equation, do we replace the series, with $S = \lim_{N \to \infty} S_N = 1$ as the sum instead of 1 (as I thought) ? – TheoryQuest1 Dec 19 '18 at 12:53
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    @TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you – caverac Dec 19 '18 at 12:56