I wish to find the convergence radius of the power series:


I found the radius is $-e-2\lt x \lt e-2$ and this was also the answer in the book.

My problem is showing the power series does not converge at the edges.

For $x=e-2$ I used the ratio test and got a limit bigger than one.

For $x=-e-2$ I get an alternating series. how do I show it does not converge?

  • 2
    $\begingroup$ Minor nitpick: the interval (or domain) of convergence is $(-e-2, e-2)$ (plus or minus the endpoints, which is what you seem to be trying to work out). The radius of convergence is half the length of this interval, i.e. $e$. $\endgroup$ – Xander Henderson Aug 8 '17 at 15:39

A useful inequality:

$$n!>\left(\frac ne\right)^n$$

which should help you show that it fails the term test at the boundaries.

The derivation of this follows by taking the log of both sides:


By the definition of the factorial, some log rules, and a Riemann sum,

$$\ln(n!)=\sum_{k=1}^n\ln(k)=\int_0^n\ln\lceil x\rceil~\mathrm dx>\int_0^n\ln(x)~\mathrm dx=n\ln(n)-n$$

  • $\begingroup$ thank you for answering. also another question. I tried checking the web but have not find a sure answer. is there does the limit $lim_{n\rightarrow \infty} (n^n)^{\frac{1}{n}}$ exists? (sorry if it's a bit off topic). $\endgroup$ – segevp Aug 8 '17 at 15:44
  • $\begingroup$ @segevp Note that $(n^n)^{1/n}=n$. Now take the limit. $\endgroup$ – Simply Beautiful Art Aug 8 '17 at 15:46
  • $\begingroup$ sorry >< i meant $lim_{n\rightarrow \infty} (n!)^{\frac{1}{n}}$ $\endgroup$ – segevp Aug 8 '17 at 15:48
  • $\begingroup$ @segevp If you use my answer, you will find that:$$(n!)^{1/n}>\frac ne$$Now take the limit $\ddot\smile$ $\endgroup$ – Simply Beautiful Art Aug 8 '17 at 15:49

Stirling's Formula $$ n!\sim\sqrt{2\pi n}\,\frac{n^n}{e^n} $$ says that $$ \frac{n!}{n^n}\sim\frac{\sqrt{2\pi n}}{e^n} $$ Thus, at the edges the absolute value of the terms are $$ |a_n|=\frac{e^nn!}{n^n}\sim\sqrt{2\pi n} $$ which means that the terms don't go to $0$ at the edges.

A Derivation of Stirling's Formula

First $$ \begin{align} \int_{k-1/2}^{k+1/2}\log(x)\,\mathrm{d}x &=\int_k^{k+1/2}\log(x)\,\mathrm{d}x+\int_{k-1/2}^k\log(x)\,\mathrm{d}x\\ &=\int_0^{1/2}\log(k+t)\,\mathrm{d}x+\int_0^{1/2}\log(k-t)\,\mathrm{d}t\\ &=\int_0^{1/2}\log\left(k^2-t^2\right)\,\mathrm{d}t\\ &=\log(k)+\int_0^{1/2}\log\left(1-\frac{t^2}{k^2}\right)\,\mathrm{d}t\tag{1} \end{align} $$ Euler's Product Formula for $\sin(\pi t)$ easily leads to $$ \sum_{k=1}^\infty\log\left(1-\frac{t^2}{k^2}\right)=\log\left(\frac{\sin(\pi t)}{\pi t}\right)\tag{2} $$ Using symmetry and a double angle formula, we get $$ \begin{align} \int_0^{1/2}\log(\sin(\pi t))\,\mathrm{d}t &=\frac12\int_0^1\log(\sin(\pi t))\,\mathrm{d}t\\ &=\int_0^{1/2}\log(\sin(2\pi t))\,\mathrm{d}t\\ &=\frac12\log(2)+\int_0^{1/2}\log(\sin(\pi t))\,\mathrm{d}t+\int_0^{1/2}\log(\cos(\pi t))\,\mathrm{d}t\\ &=\frac12\log(2)+2\int_0^{1/2}\log(\sin(\pi t))\,\mathrm{d}t\\[3pt] &=-\frac12\log(2)\tag{3} \end{align} $$ Using $(1)$, $(2)$, $(3)$, and $\log\left(n+\tfrac12\right)=\log(n)+\frac1{2n}+O\!\left(\frac1{n^2}\right)$, we get $$ \begin{align} \log(n!) &=\sum_{k=1}^n\log(k)\\ &=\int_{1/2}^{n+1/2}\log(x)\,\mathrm{d}x-\sum_{k=1}^n\int_0^{1/2}\log\left(1-\frac{t^2}{k^2}\right)\,\mathrm{d}t\\ &=\int_{1/2}^{n+1/2}\log(x)\,\mathrm{d}x-\int_0^{1/2}\log\left(\frac{\sin(\pi t)}{\pi t}\right)\,\mathrm{d}t+\sum_{k=n+1}^\infty \int_0^{1/2}\log\left(1-\frac{t^2}{k^2}\right)\,\mathrm{d}t\\ &=\int_0^{n+1/2}\log(x)\,\mathrm{d}x+\tfrac12\log(\pi)-\int_0^{1/2}\log(\sin(\pi t))\,\mathrm{d}t+\sum_{k=n+1}^\infty O\!\left(\frac1{k^2}\right)\\[6pt] &=\left(n+\tfrac12\right)\log\left(n+\tfrac12\right)-\left(n+\tfrac12\right)+\tfrac12\log(\pi)+\tfrac12\log(2)+O\!\left(\tfrac1n\right)\\[12pt] &=\left(n+\tfrac12\right)\log(n)-n+\tfrac12\log(2\pi)+O\!\left(\tfrac1n\right)\\[12pt] &=n\log(n)-n+\tfrac12\log(2\pi n)+O\!\left(\tfrac1n\right)\tag{4} \end{align} $$ Equation $(4)$ is equivalent to $$ n!=\sqrt{2\pi n}\,\frac{n^n}{e^n}\left(1+O\!\left(\frac1n\right)\right)\tag{5} $$


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