Series a_n converges and a_n>=0. Does series e^{a_n}-1 converge as well?
I suppose sum {e^{a_n}} < e^{sum{a_n}} => the sequence of partial sums converges and the series should converge. But lim e^{a_n} /= 0!What is wrong ?
The same question about convergence for series n*a_{n^2}.
Thanks for any help=)
Is there any more info about the second one? If you have some extra restrictions on the set you should say. I have come up with a solution, but it only applies if you have a few more restrictions, report back if this is the exact question.
Consider , we know that , now ASSUMING that and we have that this series and the integral share convergence. Now let then we have that
, now since we are given that converges so does
I do not have time now to do check this out thourougly but consider that is the Ratio test for this series.
Use Cauchy's Condensation test. A series satisfying the same restrictions (save continuity) of the integral test converges iff converges, so now consider . Use your laws of logarithims and this becomes the Harmonic Series, thus both this and your question series diverge by CC's test.
A series does not converge unless the sequence of numbers converges to 0. In particular, cannot converge if does not converge to 1 which is the same as saying that converges to 0.
Of course, if does converge to 0, it does not follow that [tex]\sum e^{a_n}-1[tex] converges.