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A series limit

Evaluate the limit:

    \[\ell = \lim_{x \rightarrow +\infty}\left(\sum_{n=1}^\infty\left(\frac xn\right)^n\right)^{1/x}\]

Solution

It is quite known that

    \[\left(1+\frac{1}{n} \right)^n \leq  e \leq\left(1+\frac{1}{n} \right)^{n+1}\]

Thus,

    \begin{align*} \frac{x}{(n+1)e} &\leq \frac{x}{n+1}\left(\frac{n}{n+1}\right)^n \\ &=\frac{\left(\frac x{n+1}\right)^{n+1}}{\left(\frac xn\right)^n} \\ &=\frac{x}{n}\left(\frac{n}{n+1}\right)^{n+1}\\ &\leq \frac{x}{ne} \end{align*}

Therefore, by induction , for n \geq 1,

    \[\frac{e}{n!}\left(\frac {x}{e} \right)^n \leq \left(\frac {x}{n} \right)^n \leq \frac{x}{(n-1)!}\left(\frac {x}{e} \right)^{n-1}\]

Summing the last equation we get:

    \[e\left(e^{x/e}-1\right) \leq \sum_{n=1}^\infty\left(\frac xn\right)^n \leq xe^{x/e}\]

and raising the last equation to the 1/x power , we get:

    \[\left(e\left(1-e^{-x/e}\right)\right)^{1/x}e^{1/e}\leq \left(\sum_{n=1}^\infty\left(\frac {x}{n} \right)^n\right)^{1/x}\leq x^{1/x}e^{1/e}\]

Thus, by the squeeze theorem the limit is equal to e^{1/e}.

 

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