Convergence of series for specific values of $\lambda$.

Question

Let $\lambda$ be a positive real number. For which values of $\lambda$ does the following series converge?

$$\sum_{n=1}^\infty \frac{n^{-\lambda}}{1+\lambda^{-n}}$$

I can see that the series diverges for $\lambda = 1$ (since this gives $\frac{1}{2}\times$ harmonic series). But I don't know how to prove/show if it diverges or converges for other values of $\lambda$.

Answer 1

Try the ratio test: \begin{align*} \lim_{n\rightarrow\infty}\frac{(n+1)^{-\lambda}}{1+\lambda^{-n-1}}\frac{1+\lambda^{-n}}{n^{-\lambda}}&=\lim_{n\rightarrow\infty}(1+1/n)^{-\lambda}\frac{1+\lambda^{-n}}{1+\lambda^{-n-1}}\\ &=\lim_{n\rightarrow\infty}(1+1/n)^{-\lambda}\frac{\lambda^{n+1}+\lambda}{\lambda^{n+1}+1}\\ &=\lambda\quad\text{for }\lambda<1, \end{align*} and $1$ otherwise. So, we have convergence if $\lambda<1$. For $\lambda>1$, just use the comparison test: \begin{equation*} \frac{1}{1+\lambda^{-n}}\leq 1, \end{equation*} therefore, \begin{equation*} \frac{n^{-\lambda}}{1+\lambda^{-n}}\leq n^{-\lambda}, \end{equation*} which converges for all $\lambda>1$.