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?

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

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

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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:

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

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ln 101

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

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