# Math Help - Stuck with a series problem!

1. ## Stuck with a series problem!

The series is $cosh(n)/n!$.

I attempted the ratio test with lim n→∞ of:
$\frac{\frac{\cosh(n+1)}{(n+1)!}}{\frac{\cosh(n)}{n !}}=\frac{\cosh(n+1)}{cosh(n)}(n+1)=\frac{e^(n+1)+ e^(-n-1)}{e^(n)+e^(-n)}(n+1)$

and for me this means that L>1 if evaluated when lim n→∞, and so not convergent at all. However, the problem clearly states to determine whether the series converges absolutely or conditionally - hence why my conclusion obviously has to be wrong, and I am making some sort of mistake. I hope you guys can help me see what I'm doing wrong here.

2. Shouldn't you have (n + 1) in the denominator?
p.s. use e^{.......} for exponents with more than one character.

3. Originally Posted by endreoc
The series is $cosh(n)/n!$.

I attempted the ratio test with lim n→∞ of:
$\frac{\frac{\cosh(n+1)}{(n+1)!}}{\frac{\cosh(n)}{n !}}=\frac{\cosh(n+1)}{cosh(n)}(n+1)=\frac{e^(n+1)+ e^(-n-1)}{e^(n)+e^(-n)}(n+1)$

and for me this means that L>1 if evaluated when lim n→∞, and so not convergent at all. However, the problem clearly states to determine whether the series converges absolutely or conditionally - hence why my conclusion obviously has to be wrong, and I am making some sort of mistake. I hope you guys can help me see what I'm doing wrong here.

All the terms for the cosh should be positive, no? I don't know why it would be asking you about absolute convergence.

4. Originally Posted by endreoc
The series is $cosh(n)/n!$.

I attempted the ratio test with lim n→∞ of:
$\frac{\frac{\cosh(n+1)}{(n+1)!}}{\frac{\cosh(n)}{n !}}=\frac{\cosh(n+1)}{cosh(n)}(n+1)=\frac{e^(n+1)+ e^(-n-1)}{e^(n)+e^(-n)}(n+1)$

and for me this means that L>1 if evaluated when lim n→∞, and so not convergent at all. However, the problem clearly states to determine whether the series converges absolutely or conditionally - hence why my conclusion obviously has to be wrong, and I am making some sort of mistake. I hope you guys can help me see what I'm doing wrong here.

First step is to verify if $\lim_{n \rightarrow \infty} \frac{\cosh n}{n!}=0$ ... this scope is positively achieved remembering that is...

$\ln \cosh n \sim n - \ln 2$

$\ln n! \sim \frac{\ln n}{2} + \frac{\ln 2 \pi}{2} + n \ln n - n$ (1)

The second step [ratio test] is easy if You take into account that is...

$\cosh (n+1) = \cosh n\ \cosh 1 + \sinh n \sinh 1 \implies$

$\implies \frac{\cosh (n+1)}{\cosh n} = \cosh 1 + \sinh 1 \tanh n$ (2)

... and of course that the term $n+1$ is in the denominator...

Kind regards

$\chi$ $\sigma$