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Doubt - used inequality, $1/(n^2 \log n) < 1/ n^2$ will be true only for $n > 10$ as $\log n < 1$ for $1 < n < 10 $. but here summation is running from $n= 2$ to infinite please check whether the solution which is arrived is correct with the correct procedure.

Bernard
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Abhishek Verma
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    Is it the decimal log? – Bernard Nov 25 '19 at 19:50
  • yes it is log base 10 – Abhishek Verma Nov 25 '19 at 19:53
  • What is imortant for convergence or divergence is the asymptotic behaviour. You may modify a finite number of terms in a series, it won't modify its convergence or divergence. – Bernard Nov 25 '19 at 19:56
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    In the title you are referring to LCT but it seems you are using direct comparison test. Can you clarify this point? – user Nov 25 '19 at 19:57
  • well in one of my book, there are two types are given under the title LCT, type 1 - as i have done above and type 2 - lim (Un/Vn) - L > 0 – Abhishek Verma Nov 25 '19 at 20:09
  • and I did not used type 2 because lim Un/Vn tends to 0 as n tends to infinity , but to use the properties (if Vn converges than Un too )limit should not be equal to 0 that's why I was confused whether to use that or not – Abhishek Verma Nov 25 '19 at 20:11

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Your idea is right, we have that

$$\frac{\frac1{n^2 \log n}}{\frac1{n^2}}=\frac1{\log n} \to 0$$

and therefore by LCT the given series converges.

Note that the initial values are not relevant here since we are interested to the behaviour for $n$ larg and also the inequality $1/(n^2 \log n) < 1/ n^2$ is not an issue when we use LCT.

user
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  • In this case 1/log n is strictly decreasing sequence and as n tends to infinite, 1/log n tend to 0 , but to use lct properties isn't it should not be equal to 0, example to say if Vn converges implies Un converges – Abhishek Verma Nov 25 '19 at 20:07
  • @AbhishekVerma This is not true, for the convergent case the LCT also works when the limit is equal to $0$. Refer also to here. I don't know why this point is often not clarified. – user Nov 25 '19 at 20:12
  • that's new, thanks alot – Abhishek Verma Nov 25 '19 at 20:14
  • @AbhishekVerma You are welcome! – user Nov 25 '19 at 20:17
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When testing for convergence, any initial part of the sum can be ignored, since it does not affect the result.

In your case, the initial 10 (or 100 or $10^{10}$) can be ignored.

marty cohen
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My guess is that, here, $\log$ is the natural logarithm. Anyway, it follows from the comparison test that the series $\displaystyle\sum_{n=3}^\infty\frac1{n^2\log n}$. Therefore, the series $\displaystyle\sum_{n=2}^\infty\frac1{n^2\log n}$ converges too.

José Carlos Santos
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