Wolfram is not computing it properly. What is the exact value of $$\sum_{n=1}^\infty\frac{(-1)^{n(n+1)/2}}{n}?$$ How to avoid imaginary $i$ coming from the exponent?
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Could you edit your post ? The formula is not clear to me. Where is $i$ ? – Claude Leibovici Mar 25 '14 at 09:33
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$n(n+1)/2$ is always an integer and there will be no $i$. – gammatester Mar 25 '14 at 09:36
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often 1/2 in the exponents takes precedence from the companion exponents to whom it should multiply. But wolfram compute / interprete as separate. So -1^(1/2)=i is interpreted. – waqar ahmad Mar 25 '14 at 09:42
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I think wolfram is interpreting it as $(-1)^{\left(\frac{n(n+1)}{2}\right)/n}$ Add more parenthesis. – Guy Mar 25 '14 at 09:47
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Weird. http://www.wolframalpha.com/input/?i=sum+(+(-1)%5E(n(n%2B1)%2F2))%2Fn+from+n%3D1+to+infinity this still has $i$ – Guy Mar 25 '14 at 09:48
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@Sabyasachi. What you see is the numerical value. Look at the coefficient of $i$. Cheers. – Claude Leibovici Mar 25 '14 at 09:59
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@ClaudeLeibovici I know. but why would an $i$ come in the first place. The individual terms are all real right? – Guy Mar 25 '14 at 10:06
4 Answers
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The series looks like
$$-\frac11-\frac12+\frac13+\frac14-\frac15-\frac16+\frac17+\frac18-\cdots$$
which may be reduced to
$$-\left (1-\frac13+\frac15-\cdots \right ) - \frac12 \left (1-\frac12+\frac13-\cdots \right )$$
which reduces to
$$-\frac{\pi}{4} - \frac12 \log{2}$$
Ron Gordon
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Hint: $(-1)^{n(n+1)/2}$ will take two times the value $-1$, two times the value $1$ and so on.
Writing this as two sums should return $\quad-\dfrac{\log 2}2-\dfrac{\pi}4$.
Raymond Manzoni
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Powers of $-1$ work modulo $2$, so you need to compute $$n(n+1)/2\mod 2$$ This thing has a period $4$: $$n=\{1,1,0,0,1,1,0,0,1,1,0,0,\ldots\}\mod 2$$
You can split into two sums and evaluate them separately.
orion
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Hint: Split the sum into four parts $$ \sum_{n=1}^\infty\frac{(-1)^{\frac{n(n+1)}{2}}}{n}= \sum_{n=0}^\infty\Big(-\frac{1}{4n+1} -\frac{1}{4n+2} +\frac{1}{4n+3} +\frac{1}{4n+4}\Big) $$ Then combine two parts with different signs. The partial sums up to $m$ can be expressed with the digamma function $\psi$ and the asymptotic series is $$-\frac{1}{2}\ln(2)-\frac{1}{4}\pi+ \frac{1}{4m}-\frac{9}{32m^2}+O(m^{-3})$$
gammatester
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You can't split the infinite sum into four parts, right? Each of this sum are divergent to $+\infty$. – user37238 Mar 25 '14 at 09:57
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@user37238: Thanks, splitting is OK but I cannot have four separate infinite sums. Corrected. – gammatester Mar 25 '14 at 10:03