# The result of this sum

What is the result of this sum

$$\displaystyle \sum_{n=1}^\infty (-1)^n \cdot n^2 \;?$$

And I have also tried the root test, but it gives me $$1,$$ so I can't use it, and also the alternating test is not useful.

• Welcome to Mathematics Stack Exchange. Please use MathJax. It diverges Feb 26, 2020 at 16:09
• @J.W.Tanner thank you for answering but according to what theorem or what test ? Feb 26, 2020 at 16:11
• A necessary (but not sufficient) condition for an infinite sum to be convergent is that the sequence of its general terms (a_n) converges to zero. This is not the case. Feb 26, 2020 at 16:14

$$\sum\limits_{n=1}^\infty (-1)^n n^2=\sum\limits_{k=1}^\infty -(2k-1)^2+(2k)^2=\sum\limits_{k=1}^\infty(4k-1).$$

Can you take it from here?

• I didn't understand how did you assume that n =2k and how did you ignore the -1?? Feb 26, 2020 at 16:17
• I replaced odd $n$ with $2k-1$ and even $n$ with $2k$ Feb 26, 2020 at 16:20
• And what's about the (-1)^n ???? Feb 26, 2020 at 16:22
• $(-1)^{2k-1}=-1$ and $(-1)^{2k}=1$ Feb 26, 2020 at 16:22
• @YyyUuu in case it still isn't clear what JWTanner did., rather than looking at $-1+4-9+16-25+36-49+64\pm\dots$ he looked at $(-1+4)+(-9+16)+(-25+36)+(-49+64)+\dots$, having grouped and summed the sums of consecutive terms (which that the result diverges implies the original diverges as well. one should be careful here however since if having done this causes the manipulated sum to converge this doesn't imply the original converges... take $\sum\limits_{n=1}^\infty (-1)^n$ for instance) Feb 26, 2020 at 16:38

The series diverges, by the Divergence Test (if $$\displaystyle\lim_{n\to\infty}a_n\neq0$$, then $$\displaystyle\sum_{n=1}^\infty a_n$$ diverges).

Let $$a_n=(-1)^nn^2$$ and calculate $$\displaystyle\lim_{n\to\infty}a_n$$: $$\displaystyle\lim_{n\to\infty}a_n$$ $$=\displaystyle\lim_{n\to\infty}(-1)^nn^2$$ $$=\displaystyle\lim_{n\to\infty}(-1)^n\displaystyle\lim_{n\to\infty}n^2$$

We see that neither limit converges to a value. $$\displaystyle\lim_{n\to\infty}(-1)^n$$ oscillates between $$-1$$ and $$1$$; and $$\displaystyle\lim_{n\to\infty}n^2$$ goes to infinity. Clearly $$\displaystyle\lim_{n\to\infty}a_n\neq0$$, and so the series diverges.