many struggle to solve this infinite series

[logistic_guy]

Solve.

$\displaystyle \sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n}$

 

[khansaheb]

Solve.

$\displaystyle \sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n}$

you would be surprised if i told you this

\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n} = \ln 2

www.freemathhelp.com

 

[logistic_guy]

you would be surprised if i told you this

\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n} = \ln 2

www.freemathhelp.com

Thank you.

 

[logistic_guy]

Solve.

$\displaystyle \sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n}$

$\displaystyle \sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n} = -\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n} = -\ln 2$

A detailed solution was given here:

you would be surprised if i told you this

\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n} = \ln 2

www.freemathhelp.com

Now suppose that you have no idea what a Taylor series is and you still wanna find the value of the sum. Is there a way? Yes if it is an alternating series, you can approximate the sum within what you desire!

Say you wanna guarantee to get the sum of that alternating series within $\displaystyle 0.01$.

First step take $\displaystyle \frac{1}{n}$. Add one to $\displaystyle n$, that gives $\displaystyle \frac{1}{n + 1}$.

Now solve for $\displaystyle n$.

$\displaystyle \frac{1}{n + 1} \leq 0.01$

This gives:

$\displaystyle n \geq 99$

This means that you need at least $\displaystyle 99$ terms to get an error no more than $\displaystyle 0.01$.

Let us calculate.

$\displaystyle \sum_{n=1}^{99} (-1)^{n} \frac{1}{n} = -0.698172$

The actual sum is:

$\displaystyle \sum_{n=1}^{\infty} (-1)^{n} \frac{1}{n} = -0.693147$

This shows that:

$\displaystyle -0.698172 - 0.01 \leq -0.693147 \leq -0.698172 + 0.01$

Or

$\displaystyle -0.708172 \leq \textcolor{red}{-0.693147} \leq -0.688172$

Which guarantees that our sum is within the desired approximation with an absolute error equal to:

$\displaystyle |-0.698172 - (-0.693147)| = 0.005025 < 0.01$ (Our result meets the error requirement.)

But without Taylor series, there is no way you will know that the sum converges to exactly $\displaystyle -\ln 2$.