Use Cauchy Criterion to prove that $1+\frac{2^2}{2!}+\frac{2^3}{3!}+\cdots+\frac{2^n}{n!}$ converges

Question

I've found similar problems, but they all have ones in the numerator or something that could be eventually substituted with ones, so their solutions won't work. Here's my attempt using the following rule $|x_{n+p}-x_n|<\epsilon$.

\begin{align} \frac{2^{n+1}}{(n+1)!}+\frac{2^{n+2}}{(n+2)!}+\cdots+\frac{2^{n+p}}{(n+p)!} = \end{align}

\begin{align} \frac{2^n}{n!}\biggl(\frac{2}{n+1}+\frac{2^2}{(n+1)(n+2)}+\cdots+\frac{2^p}{(n+1)\cdots(n+p)}\biggr)< \end{align}

I replaced all factors that are $\leqslant$ $\frac{2}{3}$ with $\frac{2}{3}$ (of the fraction on the left).

\begin{align} 2·\Bigl(\frac{2}{3}\Bigr)^{n-2}\biggl(\frac{2}{n+1}+\frac{2^2}{(n+1)(n+2)}+\cdots+\frac{2^p}{(n+1)\cdots(n+p)}\biggr) < \epsilon \end{align} I'm stuck here. No idea what to do with the expression in the right parentheses. Any hints appreciated.

Answer 1

You should have $$ \frac{2^n}{n!}\biggl(\frac{2}{n+1}+\frac{2^2}{(n+1)(n+2)}+\cdots+\frac{2^p}{(n+1)\cdots(n+p)}\biggr) $$ (some of your exponents are off by one).

Also $$ \frac{2^k}{(n+1)\dotsm(n+k)}\le\frac{2^k}{n^k} $$ and therefore (for $n\ge2$) $$ \frac{2}{n+1}+\frac{2^2}{(n+1)(n+2)}+\cdots+\frac{2^p}{(n+1)\cdots(n+p)}\le \sum_{k=1}^p\frac{2^k}{n^k}=\frac{2}{n}\frac{1-\dfrac{2^p}{n^p}}{1-\dfrac{2}{n}} \le\frac{2}{n}\frac{n}{n-2}=\frac{2}{n-2} $$ Thus your expression is bounded above by $$ \frac{2^{n+1}}{(n-2)\,n!} $$ Are we able to bound this with, say, $1/n$? We need that, for $n$ sufficiently big, $$ 2^{n+1}=4\cdot 2^{n-1}\le (n-2)\,(n-1)! $$ Since $2^k\le k!$ is true for every $k\ge4$, we just need $n-2\ge4$ or $n\ge6$.

Thus, for $n\ge\max\{6,1/\varepsilon\}$, the inequality $$ \frac{2^{n+1}}{(n+1)!}+\frac{2^{n+2}}{(n+2)!}+\dots+\frac{2^{n+p}}{(n+p)!}\le\varepsilon $$ holds for every $p$.

Answer 2

Notice that$$(\forall n\in\mathbb{Z}^+):\frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}}=\frac2{n+1}.$$Therefore, if $n>2$, $\dfrac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}}\leqslant\dfrac12$. Therefore, if $m\geqslant n\geqslant2$,$$\sum_{k=n}^m\frac{2^k}{k!}\leqslant\frac{2^n}{n!}\left(1+\frac12+\cdots+\left(\frac12\right)^{m-n}\right)\leqslant\frac{2^{n+1}}{n!}.$$Can you take it from here?

Answer 3

For $n\ge 4$, $$\sum_{k=1}^p\frac{2^{n+k}}{(n+k)!}\le\frac{2^n}{n!}\sum_{k=1}^\infty\bigg(\frac{2}{n}\bigg)^k=\frac{2^n}{n!}\frac{n}{n-2}\le 2\frac{2^n}{n!}.$$This doubles one of the terms in the usual expansion of $\exp 2=\sum_{m\ge 0}\frac{2^m}{m!}$; such terms are $<\epsilon/2$ for sufficiently large $n$, because $\exp 2$ is a convergent infinite series.