Positive terms: bounded equals convergent
When every a_n >= 0, the partial sums s_N can only increase: s_{N+1} = s_N + a_{N+1} >= s_N. A monotone increasing sequence converges if and only if it is bounded above, by the monotone convergence theorem. So for non-negative series, convergence is exactly boundedness of the partial sums. This is what makes every comparison test work.
The comparison test then reads: if 0 <= a_n <= b_n for all large n and sum b_n converges, so does sum a_n (its partial sums are trapped below a bound). And if a_n >= b_n >= 0 with sum b_n divergent, then sum a_n diverges too. Squeeze from above to prove convergence; push from below to prove divergence.
When the inequality is awkward to set up by hand, the limit comparison test is friendlier: if a_n, b_n > 0 and a_n/b_n -> L with 0 < L < infinity, then sum a_n and sum b_n converge or diverge together. You only need to know how a_n behaves to leading order.
The integral test settles the p-series
The standard yardstick is the p-series sum 1/n^p. To decide it cleanly we use the integral test: if f is positive, decreasing, and continuous with f(n) = a_n, then sum a_n and the improper integral of f from 1 to infinity converge or diverge together. The picture is rectangles trapped between the curve, comparing a sum to an area.
p-series sum_{n=1}^∞ 1/n^p, decided by the integral test.
Let f(x) = 1/x^p on [1, infinity): positive, continuous, decreasing
for p > 0. Compare the series to integral_1^infinity x^(-p) dx.
Case p != 1:
integral_1^R x^(-p) dx = [ x^(1-p)/(1-p) ]_1^R
= (R^(1-p) - 1)/(1 - p).
As R -> infinity:
if p > 1, then 1-p < 0, so R^(1-p) -> 0 => integral = 1/(p-1) (finite) => CONVERGES
if p < 1, then 1-p > 0, so R^(1-p) -> infinity => DIVERGES
Case p = 1:
integral_1^R (1/x) dx = ln R -> infinity => DIVERGES
(this recovers the harmonic series)
Conclusion: sum 1/n^p converges <=> p > 1.Putting comparison to work
- Look at the leading order of a_n for large n — strip away constants and lower-order terms.
- Guess a p-series 1/n^p (or geometric r^n) that matches that leading order.
- Confirm with the limit comparison test: compute a_n / (1/n^p) -> L and check 0 < L < infinity.
- Read off the answer from the known behavior of the yardstick series.
Test sum (2n + 1)/(n^3 + 5).
Leading order: numerator ~ 2n, denominator ~ n^3,
so a_n ~ 2n / n^3 = 2/n^2. Compare with b_n = 1/n^2 (p = 2 > 1).
Limit comparison:
a_n / b_n = [ (2n+1)/(n^3+5) ] * n^2
= (2n^3 + n^2) / (n^3 + 5)
-> 2 as n -> infinity (0 < 2 < infinity).
Since sum 1/n^2 converges (p = 2 > 1),
by limit comparison sum (2n+1)/(n^3+5) CONVERGES.