Vtyjkyuk

Question. Let $a_n$ be a sequence of positive terms. Pick out the cases which imply that $sum a_n$ is convergent:

1. $lim n^3/2a_n=3/2$




2. $sum n^2 a_n^2 < infty$




3. $fraca_n+1a_n < (fracnn+1)^2$, for all $n$.

My Attempt.

1. True. The given condition implies: $lim_nto infty fraca_n1/n^3/2=3/2$. Now $sum 1/n^3/2<infty$ so does $sum a_n$.




2. True. Using Holder's inequality we have: $$sum a_n le (sum 1/n^2)^1/2~(sum n^2a_n^2)^1/2<infty$$




3. I think here Ratio test cannot help us...So how can I proceed in this option...? Thank you.

As regards 3), more generally, if $a_n_ngeq 1$ and $b_n_ngeq 1$ are sequences of positive terms such that $$fraca_n+1a_nle fracb_n+1b_n$$
and $sum_nb_n$ is convergent then also $sum_na_n$ is convergent. In fact
$$a_n=a_1prod_k=1^n-1fraca_k+1a_kleq a_1 prod_k=1^n-1fracb_k+1b_kleq fraca_1b_1, b_n$$
and convergence follows by the comparison test. In your case $b_n=1/n^2$.

As an alternative to the elegant way found in the comments, by Raabe’s test we have

$$nleft(fraca_na_n+1-1right)=nfrac2n+1n^2 to 2>1$$

Since $$fraca_n+1a_n < left(fracnn+1right)^2,$$

hence $$a_n=a_1cdotfraca_2a_1cdotfraca_3a_2cdotsfraca_na_n-1<a_1cdotleft(frac12right)^2cdotleft(frac23right)^2cdotsleft(fracn-1nright)^2=fraca_1n^2.$$

Thus $$sum^infty a_nleqsum^inftyfraca_1n^2=a_1cdotsum^inftyfrac1n^2=fracpi^2a_16<infty,$$which is convergent.