# determin whether the given series converges or diverges *Sum (upper^infinity, lower n=1) (1 + 1/n)^2

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### 2 Answers

Given the series `sum_(n=1)^(oo) (1+1/n)^2` , determine whether it converges or diverges.

The ratio test is inconclusive as `lim_(n->oo)(a_(n+1))/a_n=1` .

But we can use the `"n"^(th)` term test: If the sequence `{a_n}` does not converge to 0, then the series `sum a_n` diverges.

Consider the `"n"^(th)` term of `(1+1/n)^2` : `(1+1/n)^2=1+2/n+1/n^2!=0`

**Therefore, the series diverges.**

**Sources:**

You should tell if the series is either convergent, or divergent, after performing ratio test such that:

`lim_(k->oo) (a_(k+1))/(a_k) = L`

`lim_(k->oo) ((1 + 1/(n+1))^2)/((1 + 1/n)^2)`

`lim_(k->oo) (((n+2)/(n+1))^2)/(((n+1)/n)^2)`

`lim_(k->oo) (((n+2)/(n+1))^2)/(((n+1)/n)^2) = lim_(k->oo) (n^2(n^2 + 4n + 4))/((n^2 + 2n + 1)(n^2 + 2n + 1))`

`lim_(k->oo) ((1 + 1/(n+1))^2)/((1 + 1/n)^2) = lim_(k->oo) (n^4 + 4n^3 + 4n^2)/(n^4 + 4n^3 + 5n^2 + 1)`

`lim_(k->oo) ((1 + 1/(n+1))^2)/((1 + 1/n)^2) = 1`

**Since the limit L = 1, hence the test is inconclusive and you cannot tell if the series converges or diverges.**

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