# How can I find the sum of this series?

I was just thinking how I could find the sum of this series: $$\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 4 } +...+\frac { 1 }{ 100 }$$ Could someone help me?

Note by Shashank Rammoorthy
2 years, 11 months ago

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wikipedia provides a method which gives approximately the exact value.

$$\large{\displaystyle \sum_{n=1}^{k} \frac{1}{n}=\ln(k)+\gamma+\frac{1}{2k}}$$ where $$\gamma=0.5772$$ and is the euler mascheroni constant.

the required sum is $$\large{\displaystyle \sum_{n=1}^{100} \frac{1}{n}-1=\ln(100)+0.5772+0.005-1=\boxed{4.187}}$$

- 2 years, 11 months ago

Actual answer correct to $$13$$ decimal places: $$\large{\boxed{4.1873775176396}}$$. :D :D

- 2 years, 11 months ago

To find approximate values, check Tanishq's comment.

There's no other way to find exact values except using Calculators! Else, you can do this:

$\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \ldots + \frac{1}{100} = \dfrac{\left( \frac{100!}{2} + \frac{100!}{3} + \frac{100!}{4} + \ldots + \frac{100!}{99} + \frac{100!}{100} \right) }{100!} = \dfrac{11677821270331852073640165685691639305439}{2788815009188499086581352357412492142272}$

Fact:

$\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \ldots (\text{ upto }\infty) = \infty$

- 2 years, 11 months ago

ln 101

- 2 years, 11 months ago

To know how that is an approximation, consider: $\int _{ 1 }^{ n }{ \frac { 1 }{ x } } dx$

- 2 years, 11 months ago