arithmetic progression limit $\lim_{n\to\infty} \sum_{k=1}^n\frac1{a_ka_{k+1}}$

Question

Evaluate $\lim_{n\to\infty} \sum_{k=1}^n\frac1{a_ka_{k+1}}$.

I tried to explicit in function of $a_1$ and $r$ all this expression and got:

$$\lim_{n\to\infty} \sum_{k=1}^n\frac 1{a_1^2+a_1r(2k+1)+k(k+1)r^2}$$

And it's not really helping. Any other ideas/hints?

Answer 1

Hint:

If $r$ is the ratio of the arithmetic progression, one has $$\frac1{a_k}-\frac1{a_{k+1}}=\frac{a_{k+1}-a_k}{a_ka_{k+1}}=\frac{r}{a_ka_{k+1}}$$ so we obtain a telescoping sum: \begin{align} \sum_{k=1}^n\frac1{a_ka_{k+1}}&=\frac1r\sum_{k=1}^n\biggl(\frac1{a_k}-\frac1{a_{k+1}}\biggr)=\frac 1r\biggl(\frac1{a_1}-\frac1{a_{n+1}}\biggr)=\frac 1r\frac{a_{n+1}-a_1}{a_1a_{n+1}}\\ &=\frac{(n+1)r}{ra_1(a_1+(n+1)r)}=\frac{n+1}{a_1(a_1+(n+1)r)}\\ &\sim_\infty\frac{n+1}{a_1(n+1)r}=\frac1{a_1r}. \end{align}