Verified answer

Since {1/(n + sqrt(n))} is decreasing sequence which converges to 0, the series converges by the Alternating Series Test.

However, the series converges conditionally, because Σ(n=1 to ∞) 1/(n + sqrt(n)) diverges.

To show this, I'll use the Comparison Test:

Since 1/(n + sqrt(n)) > 1/(n + n) = (1/2) * 1/n for all n = 1, 2, ..., and Σ(n=1 to ∞) 1/(2n) diverges,

being a constant multiple of the harmonic series, the claim now follows.

I hope this helps!