Convergence and absolute convergence

Could anyone please help me with this problem?

Study convergence and absolute convergence

1). ∑(k=1→∞)(-1)^k/√(k(k+1))

2). ∑(k=1→∞) (5^k/k)*(x-2)^k

Thanks a lot for looking and helping!

1) The first series converges by the Alternating Series Test, because {1/√(k(k+1))} is a decreasing sequence which converges to 0.

However, the series converges conditionally, because ∑(k=1→∞) 1/√(k(k+1)) diverges:
Since 1/√(k(k+1)) < 1/√(k(k+0)) = 1/k, and ∑(k=1→∞) 1/k is the divergent harmonic series,
the series in question diverges by the Comparison Test.

2) Using the Ratio Test:
r = lim(k→∞) |[(5^(k+1) / (k+1))*(x-2)^(k+1)] / [(5^k/k) (x-2)^k]|
..= 5|x - 2| * lim(k→∞) k/(k+1)
..= 5|x - 2|.

So, the series converges absolutely for 5|x - 2| < 1 ==> |x - 2| < 1/5.

Now, we check the endpoints.
x = 9/5 ==> ∑(k=1→∞) (-1)^k / k; conditionally convergent (as in 1 above).
x = 11/5 ==> ∑(k=1→∞) 1/k; divergent harmonic series.

So, the interval of convergence is [9/5, 11/5).

I hope this helps!