How can we prove that

$\displaystyle \lim_{n\to\infty} \sum_{k=0}^{n} \frac{{n \choose k}}{n^{k}(k+3)} = e-2$ ?

The only thing that came to my mind was writing out a few terms to see where that could get me, but I'm not able to make much progres. Any help in the form of hints or a solution would be appreciated.


Using $\displaystyle\frac{1}{k+3}=\int_0^1 t^{k+2}\ dt$, you get $$\sum_{k=0}^{n}\frac{\binom{n}{k}}{n^k(k+3)}=\int_0^1 t^2\sum_{k=0}^{n}\binom{n}{k}\frac{t^k}{n^k}\ dt=\int_0^1 t^2\Big(1+\frac{t}{n}\Big)^n\ dt,$$ and the $n\to\infty$ limit of this is $\displaystyle\int_0^1 t^2 e^t\ dt$ which has the expected value.

  • $\begingroup$ I don't get how the second integral simplifies into the given expression, could you please provide a little more info? $\endgroup$ – ExtremeRaider May 4 at 6:14
  • $\begingroup$ Do you mean you are not sure why $\sum\limits_{k=0}^{n} \frac{\binom{n}{k}}{n^k(k+3)} =\int_0^1t^2\left(1+\frac{t}{n}\right)^n\, dt$? $\endgroup$ – Minus One-Twelfth May 4 at 6:26
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    $\begingroup$ @metamorphy sorry, I don't understand what you mean by that... $\endgroup$ – ExtremeRaider May 4 at 6:36
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    $\begingroup$ @NikilKumar I mean $(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$ with $a=1$ and $b=t/n$. $\endgroup$ – metamorphy May 4 at 6:39
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    $\begingroup$ @metamorphy ohhh that makes a lot more sense. Thanks! $\endgroup$ – ExtremeRaider May 4 at 6:44

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