Math · Calculus II · Worked example
Test a p-series with the integral test
Does the series of 1/n², starting at n = 1, converge? Estimate its sum from the first ten terms.
Check the conditions
f(x) = 1/x² is positive and decreasing for x ≥ 1, and f(n) = 1/n², so the integral test applies.
Evaluate the improper integral
An antiderivative is F(x) = −1/x, so the integral from 1 to b is 1 − 1/b, which approaches 1 as b grows.
Conclude
The integral is finite, so the series converges. The same test shows the harmonic series diverges: the integral of 1/x from 1 to b is ln b, which grows without bound.
Bound the sum
The first ten terms add to 1.5498. The rest of the series lies between the integrals of 1/x² from 11 to ∞ and from 10 to ∞, which are 1/11 and 1/10. So the sum is between 1.6407 and 1.6498; Euler showed it is π²/6 ≈ 1.6449.
Result
It converges; its sum lies between 1.6407 and 1.6498 (it is π²/6).
Your turn
Does the series of 1/√n converge?
Show the answer and explanation
No.
It is a p-series with p = 1/2, and p ≤ 1, so it diverges. Its terms are even larger than those of the harmonic series.
Keep exploring
In Sequences & infinite series, set the power to 1, the harmonic series. After ten terms the partial sum is already 2.93 and still climbing, and the studio classifies it as divergent.
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