Determine if the alternating series converges: `sum_(n=2)^oo (-1)^(n)sin(1/n)`



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tiburtius's profile pic

Posted on (Answer #2)

By Leibniz criterion the series `sum_(n=1)^oo (-1)^n a_n` if

`a_1geq a_2geq a_3geq cdots`                                                         (1)


`lim_(n->oo)a_n=0`                                                                (2)

Let's now apply Leibniz criterion to our problem. In our case `a_n=sin(1/n)`.

Since `pi/2geq 1/n geq0`  for `n in NN` and sinus is monotonically decreasing function over `[pi/2,0]` ` `  we have condition (1). 

Let's now check limit


So both conditions required by Leibniz criterion are met and thus the series converges.

pramodpandey's profile pic

Posted on (Answer #3)

Let us define the series as

`sum_(n=1)^ooa_n`  ,where


(i)ignoring the sign

`a_(n+1)<a_n`   for all n





Therefore series `sum_(n=1)^oo(-1)^nsin(1/n)`   will converge.

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