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

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

`a_1geq a_2geq a_3geq cdots` **(1)**

and

`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

`lim_(n->oo)sin(1/n)=sin(lim_(n->oo)1/n)=sin0=0`

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

Let us define the series as

`sum_(n=1)^ooa_n` ,where

`a_n=(-1)^nsin(1/n)`

(i)ignoring the sign

`a_(n+1)<a_n` for all n

also

`sin(1/n)<=1/n`

Thus

`lim_(n->oo)sin(1/n)=0`

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