Looking to prove that the following series converges conditionally ∑n=1∞(−1)n+1(1+n)1nn

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Looking to prove that the following series converges conditionally  ∑n=1∞(−1)n+1(1+n)1nn

rodclassique4r · Accepted Answer

Take ∑n=1∞(−1)n+1n⋅1+nn Now, ∑n=1∞(−1)n+1n converges, whereas bn:=1+nn is monotone and bounded, thus by Abel&#039;s Test we have convergence Added on request: It is not absolutely convergent because 1+nnn1n=1+nn→1 and thus our series converges iff the harmonic one does (this is the limit comparison test), but the harmonic one does not converge...

reinosodairyshm · Accepted Answer

Hint: Use the squeeze theorem:  n1nn≤(1+n)1nn≤(2n)1nn

karton · Accepted Answer

By the AM-GM inequality, (n+1)1/n=(21⋅23⋅...⋅n+1n)1n≤1+Hnn≤1+log⁡nn hence ∑n=1N(−1)n+1(n+1)1/nn=∑n=1N(−1)n+1n+∑n=1N(−1)n+1[(n+1)1/n−1]n and the original series is the sum between a well-known conditionally convergent series and an absolutely convergent series (by the p-test). Since ∑n≥1(−1)n+1n is not absolutely convergent, the original series is conditionally convergent but not absolutely convergent.

Looking to prove that the following series converges conditionally \sum_{n=1}^\infty\frac{(-1)^{n+1}(1+n)^{\frac{1}{n}}}{n}

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