# Test the series for convergence or divergence. sum[ (1/n^2) + (1/n) n= 1.. infinity]This problem has been asked as an exercise in first semester analysis

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You need to apply limits to check whether the series is convergent or divergent.

`sum_(n=1)^oo``[(1/n^2) + (1/n)] ` = `lim_(n-gtoo)` `[(1/n^2) + (1/n)]```

`lim_(n-gtoo)(1/n^2) + lim_(n-gtoo)(1/n) = 0`

Since the limits are finite, therefore the series is convergent.

**The series `sum_(n=1)^oo` [(1/n^2) + (1/n)] converges.**

Each term is `a_n=(1/n^2)+(-1)^n(1/n)=1/n^2+(-1)^n n(1/n^2)=((-1)^n n+1)/n^2`

Now lets group the even and odd terms into one term.

`a_(2n)+a_(2n+1)=((-1)^(2n)(2n)+1)/(2n)^2+((-1)^(2n+1)(2n+1)+1)/(2n+1)^2`

`a_(2n)+a_(2n+1)=(2n+1)/(2n)^2+(-(2n+1)+1)/(2n+1)^2=(2n+1)/(2n)^2-(2n)/(2n+1)^2`

`a_(2n)+a_(2n+1)=((2n+1)^3-(2n)^3)/((2n)^2(2n+1)^2)`

`a_(2n)+a_(2n+1)=(8n^3+12n^2+6n+1-8n^3)/((2n)^2(2n+1)^2)=(12n^2+6n+1)/((2n)^2(2n+1)^2)`

and since `(12n^2+6n+1)/(16n^4+16n^2+4n^2)lt1/n^2` for all `ngt=1`

and `sum_(n=1)^(oo)1/n^2` converges then this series converges also.