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

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

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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