How can I find the sum of this series?

$\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 4 } +...+\frac { 1 }{ 100 }$

#Series #Finiteseries #SummationOfSeries

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Tanishq Varshney - 5 years, 10 months ago

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

Satyajit Mohanty - 5 years, 10 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:

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

Satyajit Mohanty - 5 years, 10 months ago

ln 101

Ashish Agarwal - 5 years, 10 months ago

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

Julian Poon - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$