I'm stucked at calculation this limit. Using $Z_n =\sum\limits_{n=1}^{\infty} \frac{1}{n}$ I could rewrite my series as $Z_{2n} - Z_n$ but I couldn't go any further since harmonic series diverges and I wasn't able to calculate the sum.
Asked
Active
Viewed 43 times
1
-
1I think the term in your sum should be $\dfrac{1}{N+n}$? And you meant $Z_N = \sum\limits_{n = 1}^N \frac{1}{n}$? – Daniel Fischer Jan 22 '16 at 08:43
-
What does $Z_n=\sum_{n=1}^\infty\frac1n$ mean? The right hand side does not depend on $n$, it is actually $\infty$. – Cm7F7Bb Jan 22 '16 at 08:44
-
@DanielFischer you're right edits done – Behrang Behvandi Jan 22 '16 at 09:07