many struggle to solve this infinite series

logistic_guy said:
Solve.

\(\displaystyle \sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n}\)
Click to expand...
\(\displaystyle \sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n} = -\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n} = -\ln 2\)

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 \(\displaystyle -\ln 2\).

:sneaky:😏
 
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