Saturday, 5 December 2015

summation - How find this sum suminftyn=1frac1n2n+a



Today Question



if a>14, show that
n=11n2n+a=π4a1eπ4a11eπ4a1+1




I have konw that solution:




n=11n2+1=π2e2π+1e2π112




solution:
note





eax=π2πa1π(12a+n=1acosnxnsinxn2+a2)
let a=1,x=0,then we have
n=11n2+1=π2e2π+1e2π112



But for (1) I have
n=11(n12)2+a14
Then I can't ,Thank you for your help.



Answer



Complete the square in the summand to get




n=11(n12)2+(a14)



Note that the sum is symmetric and equal to



12n=1(n12)2+(a14)



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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