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.

#Calculus #Series

Note by Kuldeep Guha Mazumder
5 years, 6 months ago

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Comments

could u pls explain me the second step

erica phillips - 3 years, 5 months 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.

Andrew Dennehy - 3 years, 4 months ago
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