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via the link below:.Q:
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
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