does the series 1/(n^2+n) converge or diverge? n=2 to n=infinity
you know that 1/n^2 converges
1/(n^2+n) is even less, so it does too.
To determine whether the series 1/(n^2+n) converges or diverges, we can analyze its behavior using the limit comparison test.
The limit comparison test states that if we have two positive series, a_n and b_n, and the limit as n approaches infinity of the ratio a_n/b_n exists and is nonzero, then the series a_n and b_n either both converge or both diverge.
Let's apply the test by comparing our series to the series 1/n^2, which is a well-known convergent series called the p-series with p = 2.
First, we calculate the limit as n approaches infinity of the ratio 1/(n^2+n) / (1/n^2) to determine if it is nonzero:
lim(n->∞) [1/(n^2+n)] / [1/n^2]
= lim(n->∞) [n^2/n^2] / (n^2+n)
= lim(n->∞) 1 / (1+1/n)
= 1 / (1+0)
= 1
Since the limit is nonzero, the limit comparison test tells us that both series either converge or diverge. Since the series 1/n^2 is known to converge, we can conclude that the series 1/(n^2+n) also converges.