Question

Sigma Notation Write the sum using sigma notation.

$\frac{1}{2\ln 2}-\frac{1}{3\ln 3}+\frac{1}{4\ln 4}-\frac{1}{5\ln 5}+\mathrm{....}+\frac{1}{100\ln 100}$.

Short answer

The sum using sigma notation is$\sum _{n=2}^{100}\frac{{\left(-1\right)}^n}{n\ln n}.$

Step-by-step solution

Step 1.  Given

The given sequence is$\frac{1}{2\ln 2}-\frac{1}{3\ln 3}+\frac{1}{4\ln 4}-\frac{1}{5\ln 5}+\mathrm{....}+\frac{1}{100\ln 100}$.

Step 2.  To determine

We have to write the sum using sigma notation.

Step 3.  Calculation

We try to write the given sequence using a general term with n.

$\begin{array}{l}\frac{1}{2\ln 2}-\frac{1}{3\ln 3}+\frac{1}{4\ln 4}-\frac{1}{5\ln 5}+\mathrm{....}+\frac{1}{100\ln 100}\\ =\frac{1}{2\ln 2}+\frac{-1}{3\ln 3}+\frac{1}{4\ln 4}+\frac{-1}{5\ln 5}+\mathrm{....}+\frac{1}{100\ln 100}\\ =\frac{{\left(-1\right)}^2}{2\ln 2}+\frac{{\left(-1\right)}^3}{3\ln 3}+\frac{{\left(-1\right)}^4}{4\ln 4}+\frac{{\left(-1\right)}^5}{5\ln 5}+\mathrm{....}+\frac{{\left(-1\right)}^{100}}{100\ln 100}\end{array}$

It is easy to see from the sequence that the nth term is $\frac{{\left(-1\right)}^n}{n\ln n}$.

And n goes from 2 to 100.

Hence, the sum is given by: $\sum _{n=2}^{100}\frac{{\left(-1\right)}^n}{n\ln n}$.