How to evaluate the sum: Sn=∑k=1nkk4+14

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How to evaluate the sum:  Sn=∑k=1nkk4+14

Frauental91 · Accepted Answer

So:  4k4+1=(2k2+2k+1)(2k2−2k+1)  hence  12k2−2k+1−12k2+2k+1=4k4k4+1=kk4+14  and we may notice that by setting p(x)=2x2−2x+1 we have p(x+1)=2x2+2x+1  In particular  ∑k=1nkk4+14=∑k=1n(1p(k)−1p(k+1))=1p(1)−1p(n+1)=1−12n2+2n+1  equals 2n2+2n2n2+2n+1 for any n≥1

dodato0n · Accepted Answer

Try to break the denominator into product of two factors:  4k4+1=(2k2)2+1+2(2k2)−2(2k2)  =(2k2+1)2−(2k)2  =(2k2+2k+1)(2k2−2k+1)  Using this we see the general term as:  Tk=12k2−2k+1−12k2+2k+1  Tk+1=12k2+2k+1−12(k+1)2+2(k+1)+1  Alternate terms cancel and the sum telescopes to:  1−12n2+2n+1

RizerMix · Accepted Answer

Here are the values of Sn=∑k=1nkk4+1/4 for n≤15: 45,1213,2425,4041,6061,8485,112113,144145,180181,220221,264265 One notices that the numerators are divisible by 4; dividing by 4, one recognizes the sequence of triangular numbers, tn=n(n+1)/2. Thus one can guess that Sn=4tn4tn+1. Then one can set out to prove this guess by induction. The induction step requires proving the identity 4tn4tn+1+4(n+1)4(n+1)4+1=4tn+14tn+1+1 or equivalently 4(n+1)4(n+1)4+1=4tn+14tn+1+1−4tn4tn+1

How to evaluate the sum: S_n=\sum_{k=1}^n\frac{k}{k^4+1/4}

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