Today Question
if a>14, show that
∞∑n=11n2−n+a=π√4a−1⋅eπ√4a−1−1eπ√4a−1+1
I have konw that solution:
∞∑n=11n2+1=π2e2π+1e2π−1−12
solution:
note
eax=π2πa−1π(12a+∞∑n=1acosnx−nsinxn2+a2)
let a=1,x=0,then we have
∞∑n=11n2+1=π2e2π+1e2π−1−12
But for (1) I have
∞∑n=11(n−12)2+a−14
Then I can't ,Thank you for your help.
Answer
Complete the square in the summand to get
∞∑n=11(n−12)2+(a−14)
Note that the sum is symmetric and equal to
12∞∑n=−∞1(n−12)2+(a−14)
so that we may use the residue theorem to evaluate the sum. I will skip the steps involved in deriving the general result, but will just state it here:
∞∑n=−∞f(n)=−∑kRes
where the z_k are poles of f, which obeys certain integrability conditions. Details may be found elsewhere within this site.
In this case,
f(z) = \frac{1}{\left ( z-\frac12\right)^2+\left (a-\frac14\right)}
which has poles z_{\pm} = \frac12 \pm i \sqrt{a-\frac14}. The sum is therefore equal to
-\frac{\pi}{2} 2 \Re{\left [\frac{\cot{\pi \left ( \frac12 + i \sqrt{a-\frac14}\right )}}{i 2 \sqrt{a-\frac14}} \right ]} = \frac{\pi}{\sqrt{4 a-1}} \tanh{\left ( \pi \sqrt{4 a-1}\right)}
which matches the result to be shown.
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