# Series of positive terms

I want to show that $\displaystyle 1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\cdots$ converges. I know that by using D'Alembert ratio test I easily show that this series converges but I am doing in this way: \begin{align*} s_{n}&=1+\frac{1}{2!}+\frac{1}{4!}+\cdots+\frac{1}{2(n-1)!}\\ &<1+\frac{1}{2}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{2n-3}}(\because \frac{1}{k!}\leq\frac{1}{2^{k-1}},\forall k\geq 2)\\ &=1+\frac{1}{2}[1+\frac{1}{2^2}+\frac{1}{2^4}+\cdots+\frac{1}{2^{2n-4}}] \end{align*}But as $n\to\infty$ the right hand side of the above equation becomes $$1+\frac{1}{2}.\frac{1}{1-\frac{1}{4}}=\frac{5}{3}.$$ Hence we have $s_n\leq \frac{5}{3}$. So the given positive term series is such that $(s_n)$ is bounded above hence convergent. Am O right or doing some mistake? Please suggest me!

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Looks right to me. –  anorton Dec 16 '12 at 3:36
What bothers you here? –  JohnD Dec 16 '12 at 3:36

Your proof is correct: the partial sums are monotonically increasing and bounded above, therefore they converge (and hence the series does as well).

Though it may exceed your current knowledge, the infinite series you have given converges to $$\frac{1 + e^2}{2e}$$

which can be shown using the Taylor series for the hyperbolic cosine (cosh $x$).

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