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My series is $$1+\frac{x}{2}+\frac{2! x^2}{3^2}+\frac{3!x^3}{4^3}+\ldots$$

My approach:

$$u_n= \frac{n! x^n}{(n+1)^n}$$

So, $$u_{n+1}= \frac{(n+1)! x^{n+1}}{(n+2)^{n+1}}$$

So, $$\lim_{n\to\infty}\frac{u_{n+1}}{u_n}=x$$

So if $x<1$, the series is convergent and divergent if $x>1$ (By D'Alembert's Ratio Test)

But my answer is given as: convergent if $0<x<e$ and divergent if $x\ge e$

What am I doing wrong and what test should I use instead?

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3 Answers 3

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By the ratio test we have

$$\left|\frac{u_{n+1}}{u_n}\right|=\frac{(n+1)!}{(n+2)^{n+1}}\frac{(n+1)^n}{n!}|x|=\left(1+\frac1{n+1}\right)^{-(n+1)}|x|\sim_\infty e^{-1}|x|$$ so the radius of convergence is $$R=e$$ so the series is divergent for $x>e$ and using Stirling approximation we can verify easily that the series diverges also for $x=e$.

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$$\begin{align}\frac{u_{n+1}}{u_n}&=\frac{\frac{(n+1)!}{(n+2)^{n+1}}}{\frac{n!}{(n+1)^n}}x\\ &=\left(\frac{n+1}{n+2}\right)^{n+1}x\\ &=\left(1-\frac{1}{n+2}\right)^{n+1}x\\ &=\left(1-\frac{1}{n+2}\right)^{n+2}\cdot \frac{1}{1-\frac{1}{n+2}}\cdot x\end{align}$$

Then, $$\lim_{n\to\infty }\frac{u_{n+1}}{u_n}=\frac{1}{e}x$$ and then,it converge for $x\in]-e,e[$ and diverge if $x>e$.

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  • $\begingroup$ you have put $]-e,e[$? $\endgroup$
    – RE60K
    Aug 27, 2014 at 16:00
  • $\begingroup$ $lim{(1-\frac{1}{n+2})}^{n+2}=e$ right? $\endgroup$
    – Diya
    Aug 27, 2014 at 16:22
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    $\begingroup$ No, it's $e^{-1}$. Indeed, $$\lim_{n\to\infty }\left(1+\frac{k}{n}\right)^n=e^k,$$ then for $k=-1$ we have $e^{-1}=\frac{1}{e}$. $\endgroup$
    – idm
    Aug 27, 2014 at 16:30
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$$\limsup_{n \to \infty}\frac{u_{n+1}}{u_n}=\limsup_{n \to \infty} \frac{\frac{(n+1)! x^{n+1}}{(n+2)^{n+1}}}{\frac{n! x^n}{(n+1)^n}}=\limsup_{n \to \infty} x \cdot (n+1)\cdot \frac{(n+1)^{n}}{(n+2)^{n+1}}=\\=\limsup_{n \to \infty} x \cdot \frac{(n+1)^{n+1}}{(n+2)^{n+1}}=\limsup_{n \to \infty}x \cdot \left(1-\frac{1}{n+2}\right)^{n+1}=x \cdot \frac{1}{e}$$

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  • $\begingroup$ Your last equality is not entirely correct as you wrote it. Actually, the result is correct, but you need more justification ! $\endgroup$
    – idm
    Aug 27, 2014 at 16:35

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