Show the convergence of with ratio and root test $\sum1/2^k+ 1/3^k$ I was making my homework and came with this problem. The exercise says that the root test can prove its convergence, which I've already done.
Since that $\lim_{n\to \infty+}$ of $\sqrt[k]{\frac{1}{2}^{k}+ \frac{1}{3}^k}$ = $\frac{1}{6}$ root test concludes that this series converges. But when I tried to do with the ratio test I've got $\frac{\frac{1}{2^{k+1}} + \frac{1}{3^{k+1}}}{\frac{1}{2^k} + \frac{1}{3^k}}$ $\dots$ developing and diving by $3^{k+1}$ the test yields $\frac{1}{2}$. But the exercise asks to show that the ratio test is inconclusive. In other words, to prove that $\limsup\geq$1 and $\liminf\leq$ 1. Kinda didn't understood how to proceed. Can someone help me or spot some mistake a made?
 A: I suppose that there is some confusion here. Indeed, the series$$\sum_{k=1}^\infty\left(\frac1{2^k}+\frac1{3^k}\right)$$converges and its convergence can be proved through the ratio test. My guess, is that the series whose convergence you ought to be studying is$$\frac12+\frac13+\frac1{2^2}+\frac1{3^2}+\frac1{2^3}+\frac1{3^3}+\cdots,\tag1$$which is the same as the one that you have mentioned, but without grouping terms in pairs. And, as far as the series $(1)$ is concerned, it is indeed true that it converges, but the ratio test is inconclusive: the superior limit is $\infty$, whereas the inferior limit is $0$.
A: Either you are misunderstanding the exercise, or there is an error in the exercise, because the ratio test for the series $$\sum_{k=0}^\infty\left(\frac1{2^k} + \frac1{3^k}\right)$$ is conclusive, because the limit of the ratio $\frac{a_{k+1}}{a_k}$ is indeed equal to $\frac12$.
A: $\frac  {2^{-{(k+1)}}+3^{-(k+1)}} {{2^{-k}+3^{-k}}}= \frac 1  2 \frac  {1+(2/3)^{-(k+1)}} {1+(2/3)3^{-k}} \to \frac 1  2$ which proves convergence.
