7
$\begingroup$

enter image description here

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?

$\endgroup$
  • 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 $-x^4/4$. $\endgroup$ – Fimpellizieri Nov 9 '19 at 5:00
0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $-$.

$\endgroup$
0
$\begingroup$

\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 $1.$

$\endgroup$
  • $\begingroup$ Your answer is not talking about the signs $\endgroup$ – user532874 Nov 9 '19 at 5:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.