# On divergence of Series

In this note I will prove that the series ${ \left( { S }_{ n } \right) }_{ n\ge 1 }$ given by ${ S }_{ n }=\sum _{ r=1 }^{ n }{ \frac { 1 }{ r } } \forall n\in N$ is divergent. If we can find out a divergent sub-series of ${ \left( { S }_{ n } \right) }_{ n\ge 1 }$, we can say that ${ \left( { S }_{ n } \right) }_{ n\ge 1 }$ diverges. For that we consider the sub-series ${ \left( { S }_{ { 2 }^{ n } } \right) }_{ n\ge 1 }$. We then observe that ${ S }_{ { 2 }^{ n } }=\sum _{ r=1 }^{ { 2 }^{ n } }{ { \frac { 1 }{ r } } } \\ =1+\frac { 1 }{ 2 } +\left( \frac { 1 }{ 3 } +\frac { 1 }{ 4 } \right) +\left( \frac { 1 }{ 5 } +\frac { 1 }{ 6 } +\frac { 1 }{ 7 } +\frac { 1 }{ 8 } \right) +...+\left( \frac { 1 }{ { 2 }^{ n-1 }+1 } +\frac { 1 }{ { 2 }^{ n-1 }+2 } +...+\frac { 1 }{ { 2 }^{ n } } \right) \\ >1+\frac { 1 }{ 2 } +\left( \frac { 1 }{ 4 } +\frac { 1 }{ 4 } \right) +\left( \frac { 1 }{ 8 } +\frac { 1 }{ 8 } +\frac { 1 }{ 8 } +\frac { 1 }{ 8 } \right) +...+\left( \frac { 1 }{ { 2 }^{ n } } +\frac { 1 }{ { 2 }^{ n } } +...\left( { 2 }^{ n-1 }\quad times \right) ...+\frac { 1 }{ { 2 }^{ n } } \right) \\ =1+\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } +...\left( n\quad times \right) ...+\frac { 1 }{ 2 } \right) \\ =1+\frac { n }{ 2 }$ Hence, as the sequence ${ \left( 1+\frac { n }{ 2 } \right) }_{ n\ge 1 }$ diverges, the sub-series ${ \left( { S }_{ { 2 }^{ n } } \right) }_{ n\ge 1 }$ and hence the series ${ \left( { S }_{ n } \right) }_{ n\ge 1 }$ diverges.

Note by Kuldeep Guha Mazumder
4 years, 2 months ago

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## Comments

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could u pls explain me the second step

- 2 years, 1 month ago

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They were doing something called the direct comparison test. Basically, they made a series where every term was the same or smaller than the terms of the original series, and showed that it diverged. Therefore, because a series smaller than the one given diverges, the bigger one must diverge.

- 2 years ago

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