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I'm stuck by 2 exercises on series..

  1. By using the convergence of $\Sigma ln(n)/(n^2)$, prove the convergence of $$[\sum_{k=1}^{n}ln(k)^2]-n\times ln(n)^2 + 2n\times ln(n) - 2n - 1/2\times(ln(n))^2 $$ IDK where the hell this thing comes from...

  2. $\lim_n \sum u_n = l$, $v_n = \dfrac{2}{n(n+1)}\times(\sum_{k=1}^n k\times u_k)$. Prove that $\sum v_n$ converges.(We don't know it's positive or monotonous).

Thanks a lot (°∀°)ﾉ

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    Can you please post one question at a time? – 2017-01-15
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    First time asking questions... I'm sorry... I will next time :) – 2017-01-15
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    Two questions with zero total input makes $___$ input by question, correct? – 2017-01-15
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    All those $\;\times\,'$ s are pretty confusing. Use \cdot\;$ if you have to remark there's a multiplication there. – 2017-01-15
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    By summation by parts, $$\sum_{k=1}^{n}\log^2(k) = n\log^2 n-\sum_{k=1}^{n-1} k\left(\log^2(k+1)-\log^2(k)\right) $$ and $k\left(\log(k+1)-\log k\right) = k \log\left(1+\frac{1}{k}\right)\sim 1-\frac{1}{2k} $. The first claim hence follows from Stirling's inequality $\sum_{k=1}^{n}\log(k)\sim \left(n-\frac{1}{2}\right)\log(n)-n+O(1)$. Summation by parts also solves the second exercise. – 2017-01-15
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    I see, I'll try the second. Thanks~ – 2017-01-16
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    &I'll put cdot next time, thanks for comment – 2017-01-16

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