Does it diverge or converge?

Calculus Level 3

Does the following series diverge or converge?

n = 1 sin ( ln ( n ) ) n . \sum _{n=1}^{\infty } \dfrac{\sin(\text{ln}(n))}{n}.


The Basics:
Let n = m a n \sum_{n=m}^\infty a_n be a formal infinite series. For any integer N m N\geqslant m , we define the N th N^{\text{th}} partial sum S N S_N of this series to be S N : = n = m N a n S_N:=\sum_{n=m}^N a_n ; of course, S N S_N is a real number. If the sequence ( S N ) n = m (S_N)_{n=m}^\infty converges to some limit L L as N N\to\infty , then we say that the infinite series n = m a n \sum_{n=m}^\infty a_n is convergent , and converges to L L ; we also write L = n = m a n L=\sum_{n=m}^\infty a_n , and say that L L is the sum of the infinite series n = m a n \sum_{n=m}^\infty a_n . If the partial sums S N S_N diverge , then we say that the infinite series n = m a n \sum_{n=m}^\infty a_n is divergent .

Diverge. Both. Converge. Neither.

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2 solutions

Tom Shen
May 25, 2014

i/n is the harmonic series which by intergral test is divergent and since divergent x cov. or div. is divergent, the series is divergent

Divergent times divergent is divergent? 1/n times 1/n is 1/n^2 which is convergent. So no luck with this explanation.

Bart Nikkelen - 6 years, 11 months ago

Apply Cauchy's Condensation test to see that the series is summation of sin(n(ln2)) from n=1 to infinity.....now obviously this diverges . If we consider the sequence of partial sums....we see that we can find two subsequences which converge to different limits.

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