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If $a_1=1$ and $a_n=n(1+a_{n-1})$ $\forall n\geq 2$, then evaluate the given limit.

$$\lim_{n\to \infty} \bigg(1+\frac{1}{a_1}\bigg)+\bigg(1+\frac{1}{a_2}\bigg)+\cdots+\bigg(1+\frac{1}{a_n}\bigg)$$

Usually such type of questions are solved by squeeze theorem or by converting them into definite integral but don't see neither working here. Could someone give me little help to proceed

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  • $\begingroup$ Did you mean to sum or multiply the terms? $\endgroup$ – robjohn Dec 31 '17 at 2:33
  • $\begingroup$ @robjohn Multiply. Had put plus sign by mistake $\endgroup$ – Mathematics Dec 31 '17 at 5:24
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Define $$b_n=\sum_{i=1}^n\left(1+\frac{1}{a_i}\right)$$

We are interested in $\lim_{n\to\infty}b_n$. First notice that we have that $a_i>0$, since the sequence starts at $1$ and is monotonic. It follows that $b_n>n$ and so the limit diverges.

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  • $\begingroup$ Sorry. Actually the terms are in multiplication. $\endgroup$ – Mathematics Dec 31 '17 at 5:25
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    $\begingroup$ @Mathematics Given the substantial reversion which fundamentally changes the problem, the best option would be to keep the old question and post what you meant to ask as a new question. $\endgroup$ – Stella Biderman Dec 31 '17 at 5:27
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A good way of looking at it is to write the limit this way:

$$\lim_{n\to\infty}\left(\Bigl(1 + 1 + 1 + \cdots \Bigr) + \left(\frac{1}{a_1} + \frac{1}{a_2} + \cdots \right)\right) = \lim_{n\to\infty}n + \lim_{n\to\infty}\left(\frac{1}{a_1} + \frac{1}{a_2} + \cdots \right).$$

Let $$x = \lim_{n\to\infty}\left(\frac{1}{a_1} + \frac{1}{a_2} + \cdots \right).$$

Then we have $$\infty + x = \infty.$$

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    $\begingroup$ This answer would be greatly improved by using MathJax formatting to improve its legibility. $\endgroup$ – Stella Biderman Dec 31 '17 at 2:24

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