

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

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!

share|cite|improve this question
Looks right to me. – apnorton 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 $x$).

share|cite|improve this answer

Yes, you are doing correct. The technique you used, i.e. upperbounding every term in the sequence by a larger term is referred to as comparison testing. Look here for more details

share|cite|improve this answer

Maybe you can compare it to exp(1). You would consider the term of your series to be a2n = 1/(2n!) , a2n+1 = 0. Obviously for each k, ak<=1/(k!) and therefore your series converges and its sum is less than exp(1).

share|cite|improve this answer

Your Answer


