$$\sum_{x=1}^\infty (e-(1+\frac{1}{x})^x)$$ I wrote a python script to add up the first few thousand terms and the series does not seem to converge, but I have no idea how to prove this efficiently.(I would prefer if the solution did not just involve using a series test and then using l'hopital's rules several times)
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2Is that supposed to be $\sum_{x=1}^\infty [e^x - (1+x^{-1})^x]$? – eyeballfrog Jul 01 '19 at 15:25
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Your sum does not converge! – Dr. Sonnhard Graubner Jul 01 '19 at 15:26
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This obviously does not converge as the RHS term converges to the finite value $e$ whereas $e^x$ is unbounded so the terms do not tend to zero and hence the sum cannot converge. – Peter Foreman Jul 01 '19 at 15:27
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1I'm wondering if this sum was intended to be $e-(1+\frac{1}{x})^x$ – Brenton Jul 01 '19 at 15:29
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Perhaps is $\displaystyle\sum e^n-(1+\frac{n}{x})^x$ for some $x$! – Nosrati Jul 01 '19 at 15:29
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I guess as a further question, is that supposed to be $\sum_{n=1}^\infty [e^{x} - (1+x/n)^n]$, a sum which also diverges but less trivially so. – eyeballfrog Jul 01 '19 at 15:30
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Something in the "similar" spirit: Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$. Not a duplicate, of course. – Jul 01 '19 at 15:31
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Sorry I clarified the question – Anirudh Jul 01 '19 at 15:33
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2@herbsteinberg I'm not the OP, but I think it is alright the way it is. The sequence $$a_n = e - \left(1 + \frac 1 n \right)^n$$ does approach $0$ as $n\to\infty$, so it is reasonable to ask if $\sum^\infty_{n=1} a_n $ converges. – User8128 Jul 01 '19 at 15:43
3 Answers
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$$\begin{align} \left(1+\frac1x\right)^x &=\exp{\left(x\ln{\left(1+\frac1x\right)}\right)}\\ &=\exp{\left(x\left(\frac1x-\frac1{2x^2}+O\left(\frac1{x^3}\right)\right)\right)}\\ &=\exp{\left(1-\frac1{2x}+O\left(\frac1{x^2}\right)\right)}\\ \end{align}$$
Using the series expansion above (valid for $x\ge1$) we have $$\begin{align} e-\left(1+\frac1x\right)^x &=e-\exp{\left(1-\frac1{2x}+O\left(\frac1{x^2}\right)\right)}\\ &=e\left(1-\exp{\left(-\frac1{2x}+O\left(\frac1{x^2}\right)\right)}\right)\\ &=e\left(1-\left(1+\left(-\frac1{2x}+O\left(\frac1{x^2}\right)\right)+O\left(\frac1{x^2}\right)\right)\right)\\ &=e\left(1-\left(1-\frac1{2x}+O\left(\frac1{x^2}\right)\right)\right)\\ &=e\left(\frac1{2x}+O\left(\frac1{x^2}\right)\right)\\ &=\frac{e}{2x}+O\left(\frac1{x^2}\right)\\ \end{align}$$ $$\sum_{x=1}^n\left(e-\left(1+\frac1x\right)^x\right)=\sum_{x=1}^n\left(\frac{e}{2x}+O\left(\frac1{x^2}\right)\right)=\frac{e}2H_n+O(1)$$ where $H_n$ is the $n$th harmonic number and $O$ denotes big-O notation. As $\lim_{n\to\infty}H_n$ is well known to diverge, then our sequence also diverges.
Peter Foreman
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1How did you arrive at that series expansion(I am not familiar with the big O notation) – Anirudh Jul 01 '19 at 15:49
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1I've edited the answer to include the workings out of the series expansion, but as for the big-O notation, I've provided a link to the Wikipedia article on the topic. – Peter Foreman Jul 01 '19 at 15:59
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$$ e - \left(1+\frac{1}{x}\right)^x = \sum_{n=0}^\infty \frac{1}{n!} - \sum_{n=0}^x \frac{x!}{n!(x-n)!}x^{-n} = \sum_{n=2}^x \frac{1}{n!}\left[1 - \frac{x!}{(x-n)! x^n}\right] + \sum_{n=x+1}^\infty \frac{1}{n!} $$ The $n=0,1$ terms drop out because they are zero. Note that every term in these sums is positive, since $x^n(x-n)! > x!$ for all $1 < n \le x$. In particular, that means that if we drop every term except $n=2$, we get $$ e - \left(1+\frac{1}{x}\right)^x > \frac{1}{2x}. $$ Thus, $$ \sum_{x=1}^\infty \left[ e - \left(1+\frac{1}{x}\right)^x\right] > \sum_{x=1}^\infty \frac{1}{2x} = \infty, $$ and the series diverges.
eyeballfrog
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Let's make use of the fact that $e\gt(1+1/N)^N$ for any positive integer $N$. Then, letting $N=2n$ and using the algebraic identity $a^n-b^n=(a-b)(a^{n-1}+\cdots+b^{n-1})$, we have
$$\begin{align} e-\left(1+{1\over n}\right)^n &\gt\left(1+{1\over2n}\right)^{2n}-\left(1+{1\over n}\right)^n\\ &=\left(\left(1+{1\over2n}\right)^2-\left(1+{1\over n}\right) \right)\left(\left(1+{1\over2n}\right)^{2(n-1)}+\cdots+\left(1+{1\over n}\right)^{n-1} \right)\\ &\gt{1\over4n^2}\left(1+\cdots+1\right)\\ &={1\over4n} \end{align}$$
and thus
$$\sum_{n=1}^\infty\left(e-\left(1+{1\over n}\right)^n\right)\gt{1\over4}\sum_{n=1}^\infty{1\over n}$$
which diverges.
It's worth noting that the algebraic steps above show that
$$\left(1+{1\over N}\right)^N \lt\left(1+{1\over2N}\right)^{2N} \lt\left(1+{1\over4N}\right)^{4N} \lt\cdots\left(1+{1\over2^kN}\right)^{2^kN}\lt\cdots$$
which proves the fact that $e\gt(1+1/N)^N$ for any positive integer $N$, since $2^kN\to\infty$ as $k\to\infty$.
Barry Cipra
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