Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
    
Looks right to me. –  anorton Dec 16 '12 at 3:36
    
What bothers you here? –  JohnD Dec 16 '12 at 3:36
add comment

3 Answers

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|improve this answer
add comment

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|improve this answer
add comment

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|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.