0
$\begingroup$

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 $\lambda$.

$\endgroup$
1
$\begingroup$

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 $\lambda>1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.