How to solve the following problem?

Let {${a_n}$} be an increasing sequence of positive real numbers such that the series $\sum_{k=1}^{\infty}{a_k}$ is divergent. Let $S_n$=$\sum_{k=1}^{n}{a_k}$ for $ n=1,2,...$ and $t_n$=$\sum_{k=2}^{n}{\frac{a_k}{S_{k-1}S_k}}$ for $n=2,3,... $

Then $lim_{n\to\infty}{t_n}$ is equal to

(a) $\frac{1}{a_1}$

(b) $0$

(c) $\frac{1}{a_1+a_2}$

(d) $a_1+a_2$


closed as off-topic by Did, Martin R, user296602, Namaste, Saad Mar 28 '18 at 1:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Martin R, Community, Namaste, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.


It would be appreciated if the homework questions policy was observed before asking this question.

Others have pointed towards different versions of this question, but following the policy of giving a hint at first and a full answer when the homework is expected to be due (for completeness of this site), here is your hint:

$$ a_k = S_k-S_{k-1} $$

So: $$ \frac{a_k}{S_k S_{k-1}}=\frac{S_k-S_{k-1}}{S_k S_{k-1}}=\frac{1}{S_{k-1}}-\frac{1}{S_k} $$

  • 1
    $\begingroup$ Exactly that hint was given at math.stackexchange.com/a/1115979. $\endgroup$ – Martin R Mar 27 '18 at 14:00
  • $\begingroup$ @MartinR Yes indeed. $\endgroup$ – Mefitico Mar 27 '18 at 14:27
  • $\begingroup$ oh! that's pretty easy! I couldn't find or think about this hint before.. Thank you @Mefitico $\endgroup$ – user511110 Mar 27 '18 at 16:31

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