# Evaluate $\int x^2\ln (1+x) \, dx$ as a power series: why is just $n$ ok?

This is a different question than the previous one I posed pertaining to the same textbook problem. I do not understand the justification in step seven for the exponent not changing. If you add $1$ to all the numbers that make up the set of $n$, then you must cancel out this effect by turning every $n$ into $n-1$ so that the series remains the exact same. You must have $(-1)^{n-1}$. What am I missing?

• 5$\begingroup$ It doesn’t make sense. The first term of the series in $(6)$ is $x^4/4$ while the first term in $(7)$ is $-x^4/4$. $\endgroup$ Nov 9, 2019 at 5:00

You’re right it’s not correct as currently stated. Since there’s a specific note about not changing the exponent of $(-1)^{n}$, the author obviously was aware of this issue. Perhaps the textbook author intended to, but forgot, to change the “$C +$” part to “$C –$” (i.e., effectively moved a factor of $-1$ outside of the summation)? This would be a minimal mistake to have made, even though the final comment doesn’t seem to completely fit this potential scenario. Also, it makes more sense, at least to me, to use $n-1$ as an exponent rather than changing the $+$ of the summation terms to a $-$.

You are right, the sign is missing.

The alternative Maclaurin expansion of $\ln (1+x)$ is: $$\ln (1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n}=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n}.$$ Note: The sum starts from $n=1$ already.

Hence: $$\int x^2\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n} dx=\int \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{n+2}}{n} dx=\\ C+\sum_{n=1}^\infty \frac{(-1)^{n+1}x^{n+3}}{n(n+3)}=C+\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n+3}}{n(n+3)}.$$

\begin{align} & \sum_{n=0}^5 x^{n+4} \quad \text{(starting with $n=0$)} \\[10pt] = {} & x^4 + x^5 + x^6 + x^7 + x^8 + x^9 \\[10pt] = {} & \sum_{n=1}^6 x^{n+3} \quad \text{(starting with $n=1$)} \end{align} “Fimpellizieri” is right: A sign change was neglected. The exponent in $(-1)^n$ should also have changed by $1.$

• $\begingroup$ Your answer is not talking about the signs $\endgroup$– user532874Nov 9, 2019 at 5:27

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