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How to calculate the error between the solution given by Wolfram Alpha and my solution?

I have an exercise to calculate the error between the solution given by Wolfram Alpha and my solution.
The exercise is this:

The sum of the first 4 terms of the series is $$\sum_{n=1}^\infty \frac{(-1)^n}{n}$$
Find the error between the series I have found (say it’s correct) and the answer given by Wolfram Alpha.

This is what I’ve done so far:

$\sum_{n=1}^\infty \frac{(-1)^n}{n} = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{n+1}$
$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}$ = $\sum_{n=0}^\infty \frac{(-1)^{n+1}}{n+2}$
Hence, by the argument principle, we have that $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}$ = $\sum_{n=0}^\infty \frac{(-1)^{n+1}}{n+2}$ = $\frac{1}{1} + \frac{1}{0}$ = $\frac{1}{0} + \frac{1}{0}$ = $-1$,
which gives us our series.
$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1} = -1$

Now, I’m a bit stuck. I feel the error is too large.

A:

The sum of the first $4$ terms of your series is
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\frac12-\frac{1}{4}\cdot3+\frac{1}{6}\cdot3-\frac{1}{8}\cdot3+\cdots,$$
and the sum of the first \$4

https://ello.co/7mijuvcontnu/post/jiq_oesuuy2lpqqqtcoxka
https://ello.co/8perryomoe_ji/post/fnqhckyip1rincaz7lhagq
https://ello.co/0riatuqui_pi/post/ylpvdw9mkk_bzqs9tozlwg
https://ello.co/cisaxsiugo/post/ob2nj8z8dxa9xl7rmfaxeq
https://ello.co/atmainga/post/nutgjitg68xbs1aguww7xa
https://ello.co/0tiamaynest-su/post/kcaq5fk70bouu2kjrm-r-w
https://ello.co/penbalocont_ge/post/mbekhtietyahw0vqosfvcw

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