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)=−∑kResz=zkπcot(πz)f(z)
where the zk are poles of f, which obeys certain integrability conditions. Details may be found elsewhere within this site.
In this case,
f(z)=1(z−12)2+(a−14)
which has poles z±=12±i√a−14. The sum is therefore equal to
−π22ℜ[cotπ(12+i√a−14)i2√a−14]=π√4a−1tanh(π√4a−1)
which matches the result to be shown.
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