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Knowing that $\lim_{n\to\infty}x_n=x$ I want to prove that $$\lim_{n\to\infty}\frac{x_1+2^kx_2+\dots+n^kx_n}{n^{k+1}}=\frac{x}{k+1}.$$

My guess is that we will use the Stolz–Cesàro theorem.

So for $$a_n=x_1+2^kx_2+\dots+n^kx_n$$and $$A_n=n^{k+1}$$ I have:

$$\frac{a_{n+1}-a_n}{A_{n+1}-A_{n}}=$$ $$\frac{x_1+2^kx_2+\dots+(n+1)^kx_{n+1}-x_1-2^kx_2-\dots-^kx_n}{(n+1)^{k+1}-n^{k+1}}=$$ $$\frac{(n+1)^kx_{n+1}}{(n+1)^{k+1}-n^{k+1}}$$

From here on, I'm sure if I continue correctly:

$$\frac{x_{n+1}}{\frac{(n+1)^{k+1}-n^{k+1}}{(n+1)^k}}=$$ $$\frac{x_{n+1}}{\frac{(n+1)^{k+1}}{(n+1)^k}-\frac{n^{k+1}}{(n+1)^k}}=$$ $$\frac{x_{n+1}}{n+1-(\frac{n}{n+1})^kn}=$$ $$\frac{x_{n+1}}{n+1-(\frac{1}{\frac{n+1}{n}})^kn}$$

And that is what I got so far. I've got the numerator correct and the denominator seems close.

Any tips on how to expand the denominator?

Paramanand Singh
  • 87,309
Shevliaskovic
  • 1,017

2 Answers2

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$$n+1-\left(1+\frac1n\right)^{-k}n=n\left[1-\left(1+\frac1n\right)^{-k}\right]+1$$

Now: with $\;x\in\Bbb R\;$ a continuous variable, and applying l'Hospital:

$$\lim_{x\to\infty}\frac{1-\left(1+\frac1x\right)^{-k}}{\frac1x}=\lim_{x\to\infty}\frac{k\left(1+\frac1x\right)^{-k-1}\left(-\frac1{x^2}\right)}{\left(-\frac1{x^2}\right)}\xrightarrow[x\to\infty]{}k$$

and now you can end with the result you wanted using arithmetic of limits.

Timbuc
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$(n+1)^{k+1}-n^{k+1}$= $n^{k+1}[(1+\frac{1}{n^{k+1}})^{k+1}-1]$

Aproximation is $n^{k+1}[1+ \binom{k+1}{1}n^{k+1}-1+o(n^{k+1})]$ and that is k+1(as you want to get)

user207950
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