Need to prove the sequence: an=1+122+132+…+1n2 converges.

Jamarion Kerr

Jamarion Kerr

Answered

2022-01-22

Need to prove the sequence:
an=1+122+132++1n2 converges.

Answer & Explanation

Gordon Stephens

Gordon Stephens

Expert

2022-01-23Added 10 answers

Notice that 2k2k(k+1)1k22k(k+1)
k=12k(k+1)=21×2+22×3+23×4+
k=12k(k+1)=2((112)+(1213)+(1314)+)
k=12k(k+1)=2(1)=2
Therefore k=11k22
Micheal Hensley

Micheal Hensley

Expert

2022-01-24Added 10 answers

an=1n2 and because an>0 we have |an|=an
First: check the necessary condition
limnnan=limn1n=0
Second: check DAlemberts ratio test
limn+|an+1an|=limn+an+1an=limn+(nn+1)2=1
Third: Because the answer of DAlemberts test is 1, you should use Raabes
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

This here should work with n1 sn=n=11n2=11+14+19+116+...+1n2 bn=n=112n1=11+12+14+18+...+12n1 bn is directly compared greater than sn: sn<bn and bn converges, because of its ratio test : 12n1+1/12n1=2n12n=12<1

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