Is the series \sum_{n=1}^\infty\frac{1}{n}(\sum_{k=1}^n\frac{1}{k}(\frac{1}{2})^{n-k})

Loretta Cannon

Loretta Cannon

Answered question

2022-02-27

Is the series
n=11n(k=1n1k(12)nk)

Answer & Explanation

cassidacc4

cassidacc4

Beginner2022-02-28Added 7 answers

If we denote
f(a,n)=k=1neakkf(x,n)=ddxf(x,n)=k=1nexk and
f(a,n)=af(x,n)dx
Then the initial sum
S=n=11n(k=1n1k(12)nk)=n=11n12nk=1n2kk
=1n12nf(ln2,n)
=n=11n12nln2f(x,n)dx
Evaluating f(x,n)
f(x,n)=k=1nexk=ex(1exn)1ex
Therefore, our sum can be presented in the form
S=n=11n12nln2ex(1exn)1exdx
=ln2ex1exdx
Tye Rhodes

Tye Rhodes

Beginner2022-03-01Added 4 answers

By the Stolz-Cesàro Theorem, we have
k=1n2kk2n+1n
So
1nk=1n1k(12)nk=1n2nk=1n2kk1n2n2n+1n=2n2
Hence, the series converge.

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