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I stumbled upon this sum:

$$\sum e^{an^2} (1-\frac{a}{n})^{n^3}$$ Wolfram Alpha tells me it is convergent but I can't find a convenient proof to use.

The $n$-th root test seems to be useful, since we get $e^{an}(1-\frac{a}{n})^{n^2}$ and apparently this tends to $e^{-\frac{a^2}{2}}<1$ as long as $a\neq 0$. I say apparently because Wolfram says so but I can't seem to prove it; passing to logarithms, I can't use the equivalent $\log(1+u)\sim u$ when $u\to 0$ since things cancel out.

In any case, my main question is: can you provide a proof for convergence that doesn't rely on the root test? My bonus question is, please help me compute the limit above.

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Main Answer

Using the power series for $\log(1-x)$, we have $$ \begin{align} n^3\log\left(1-\frac an\right) &=n^3\left[-\frac an-\frac12\frac{a^2}{n^2}-\frac13\frac{a^3}{n^3}+O\left(\frac1{n^4}\right)\right]\\ &=-an^2-\frac12a^2n-\frac13a^3+O\left(\frac1n\right) \end{align} $$ Therefore, $$ \begin{align} e^{\large an^2}\left(1-\frac an\right)^{\large n^3} &=\exp\left[-\frac12a^2n-\frac13a^3+O\left(\frac1n\right)\right]\\ &=O\left(e^{\large-\frac12a^2n}\right) \end{align} $$ which is summable for $a\ne0$.


Bonus Answer

Since $$ \begin{align} n^2\log\left(1-\frac an\right) &=n^2\left[-\frac an-\frac12\frac{a^2}{n^2}+O\left(\frac1{n^3}\right)\right]\\ &=-an-\frac12a^2+O\left(\frac1n\right) \end{align} $$ Therefore, $$ \lim_{n\to\infty}e^{an}\left(1-\frac an\right)^{\large n^2}=e^{\large-\frac12a^2} $$

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  • $\begingroup$ Thanks. For the bonus question, I should have bothered to use the Taylor development a bit further, and in fact I see that more or less the same trick worked for the series. $\endgroup$ – user46225 Dec 2 '14 at 10:16
  • $\begingroup$ How can you justify the step after the "therefore"s ? Honestly, it was always taught to me with some words along the lines "you can replace by an equivalent in a sum as long as it doesn't cancel all out", which was in fact the reason for my method not working for the computation of the limit. Besides, how do you justify doing it inside of an exponential? It isn't always true that $f\sim g$ implies $e^f \sim e^g$, but when is it? $\endgroup$ – user46225 Dec 2 '14 at 21:09
  • $\begingroup$ I've expanded the above comment into a new question, please do take a look: math.stackexchange.com/questions/1048852/… $\endgroup$ – user46225 Dec 2 '14 at 21:11
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    $\begingroup$ @user46225: The other question is a bit too broad. Let me address the situation here. The reason that I use big-O notation is that it is more precise than $\sim$. Since $$n^3\log\left(1-\frac an\right) =-an^2-\frac12a^2n-\frac13a^3+O\left(\frac1n\right)$$ we can exponentiate and multiply by $e^{an^2}$ to get $$e^{\large an^2}\left(1-\frac an\right)^{\large n^3}=\exp\left[-\frac12a^2n-\frac13a^3+O\left(\frac1n\right)\right]$$ $\endgroup$ – robjohn Dec 3 '14 at 0:20
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    $\begingroup$ @user46225: From here, we get that $$e^{\large an^2}\left(1-\frac an\right)^{\large n^3}\sim\exp\left[-\frac12a^2n-\frac13a^3\right]$$ $\endgroup$ – robjohn Dec 3 '14 at 0:47
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Our first test to try is the limit test, since that exponential looks suspiciously divergent to me. $$ \begin{align} &\lim_{n\to\infty}e^{an^2}\left(1-\frac a n\right)^{n^3}\\ =&\lim_{n\to\infty}\exp\left\{\log\left(e^{an^2}\left(1-\frac a n\right)^{n^3}\right)\right\}\\ =&\lim_{n\to\infty}\exp\left\{an^2+n^3\log\left(1-\frac a n\right)\right\} \end{align} $$ As $n$ becomes large, $\frac a n$ becomes small. We can take a series approximation to the logarithm and (after expansion) keep the dominant term: $$ \begin{align} =&\lim_{n\to\infty}\exp\left\{an^2+n^3\left(-\frac a n-\frac{a^2}{2n^2}-\frac{a^3}{3n^3}-\frac{a^4}{4n^4}-\cdots\right)\right\}\\ =&\lim_{n\to\infty}\exp\left\{an^2-an^2-\frac{a^2n}{2}-\frac{a^3}{3}-\frac{a^4}{4n}-\cdots\right\}\\ =&\lim_{n\to\infty}\exp\left\{-\frac{a^2n}{2}-\cdots\right\}\\ =e^\frac{a^2}2&\lim_{n\to\infty}e^{-n}=0 \end{align} $$ Note that the last line holds only if $a\neq0$, otherwise every term in the series is $1$. Therefore the sum diverges at $a=0$. I bet that the convergence has something to do with the fact that $\lim_{n\to\infty}\left(1+\frac 1n\right)^n=e$.

Now we can move on to the ratio test. $$ \begin{align} &\lim_{n\to\infty}\frac{e^{a(n+1)^2}\left(1-\frac a{n+1}\right)^{(n+1)^3}}{e^{an^2}\left(1-\frac a n\right)^{n^3}}\\ =&\lim_{n\to\infty}\exp\left\{a(n+1)^2-an^2+(n+1)^3\log\left(1-\frac a{n+1}\right)-n^3\log\left(1-\frac a n\right)\right\}\\ =&\lim_{n\to\infty}\exp\left\{a+2an-an^2-\frac{4+a}2an-\cdots+an^2+\frac a2an+\cdots\right\}\\ =&\lim_{n\to\infty}\exp\left\{a+2an-\frac 42an-\frac 12(2+a)a\cdots\right\}\\ =&\lim_{n\to\infty}\exp\left\{-\frac{a^2}2+\cdots\right\}=e^{-a^2/2} \end{align} $$ (I'll admit I had Mathematica calculate those series!) Since we know from before that $a\neq0$, $-\frac{a^2}2<0$ (since $a^2$ is positive for all real nonzero $a$) and therefore its exponential must be less than $1$.

Therefore by the ratio test the sum converges for all $a\neq0$ and by the limit test the sum diverges for $a=0$.

Now, for extra credit, I'll do the root test. You correctly calculated the form of the limit: $$ \begin{align} &\lim_{n\to\infty}\sqrt[n]{e^{an^2}\left(1-\frac a n\right)^{n^3}}\\ =&\lim_{n\to\infty}\exp\left\{\frac 1n\left(an^2+n^3\log\left(1-\frac a n\right)\right)\right\}\\ =&\lim_{n\to\infty}\exp\left\{an+n^2\log\left(1-\frac a n\right)\right\} \end{align} $$ However, you got stuck when the first term of the $\log$ series expansion canceled. The full series expansion (which I used above) is $$ \log(1+u)=u+\frac{u^2}2+\frac{u^3}3+\cdots+\frac{u^n}n+\cdots $$ I'll use the first three terms to evaluate our limit: $$ \begin{align} =&\lim_{n\to\infty}\exp\left\{an+n^2\left(-\frac a n-\frac{a^2}{2n^2}-\frac{a^3}{3n^3}-\cdots\right)\right\}\\ =&\lim_{n\to\infty}\exp\left\{an-an-\frac{a^2}2-\frac{a^3}{3n}-\cdots\right\}\\ =&\lim_{n\to\infty}\exp\left\{-\frac{a^2}2-\frac{a^3}{3n}-\cdots\right\} \end{align} $$ The first term canceled, but the rest of the terms stay. Note that of the remaining terms, the second has $n$ in the denominator, so in the limit it tends to $0$. In fact, since each one of the terms following has one more power of $n$ in the denominator they all tend to $0$. This leaves us with our final expression: $$ =\lim_{n\to\infty}\exp\left\{-\frac{a^2}2-\cdots\right\}=e^{-a^2/2} $$ Which, as before, is less than $1$ when $a\neq 0$, meaning that by the root test the series converges when $a\neq 0$.

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  • $\begingroup$ Very nice and thorough answer, thank you very much for taking the time. $\endgroup$ – user46225 Dec 2 '14 at 10:18

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