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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.2012-12-16
  • 0
    What bothers you here?2012-12-16

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