Thread: Find radius of convergence without using l'hopital...

$\displaystyle \sum_{k=0}^{\infty}\frac{x^{2k}}{log(1+k)}$

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Thanks

Originally Posted by Aileys.

$\displaystyle \sum_{k=0}^{\infty}\frac{x^{2k}}{log(1+k)}$

?

Thanks

put a_n = x^(2n)/log(1+n) ==> |a_(n+1)/a_n| = |x|^2*[log(1+k)/log(2+k)] --> |x|^2 < 1 <==> -1 < x < 1.

Now they must mean not to use L'H in the lim above, but we can done the following (squeeze or sandwich theorem)

log(k)/log(2k) <= log(1+k)/log(2+k) <= log(1+k)/log(1+k) = 1

We use that log(x) is an monotonically ascending function. But

log k/log 2k = 1/[log 2/log k + 1] --> 1/[0+1] = 1 and we're done.

Tonio

Very nice, thanks a lot