×

# 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, 6 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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, 6 months ago

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

- 2 years, 6 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, 6 months ago

ln 101

- 2 years, 6 months ago

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

- 2 years, 6 months ago