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Exercise 393Calculus › Single-variable calculus › Series › Convergence testsL1Added 2 Nov 2026

The comparison test in practice

#comparison test#p-series

Statement

Decide whether each series converges.

  1. ∑1n2+n\displaystyle\sum\frac1{n^2+n}
  2. ∑1n3+1\displaystyle\sum\frac1{\sqrt{n^3+1}}
  3. ∑n+1n2−n+1\displaystyle\sum\frac{n+1}{n^2-n+1}
  4. ∑sin⁡2n2n\displaystyle\sum\frac{\sin^2n}{2^n}
Hint

For positive terms, if 0≤un≤vn0\le u_n\le v_n and ∑vn\sum v_n converges, so does ∑un\sum u_n; if ∑un\sum u_n diverges, so does ∑vn\sum v_n.

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Solution

1. 1n2+n≤1n2\frac1{n^2+n}\le\frac1{n^2}, a convergent pp-series (p=2p=2): converges. (It even telescopes: 1n−1n+1\frac1n-\frac1{n+1}.)

2. 1n3+1≤1n3/2\frac1{\sqrt{n^3+1}}\le\frac1{n^{3/2}}, a convergent pp-series (p=32>1p=\frac32>1): converges.

3. For n≥2n\ge2, n2−n+1≤n2n^2-n+1\le n^2, so n+1n2−n+1≥n+1n2≥1n\frac{n+1}{n^2-n+1}\ge\frac{n+1}{n^2}\ge\frac1n. The harmonic series diverges: diverges.

4. 0≤sin⁡2n2n≤12n0\le\frac{\sin^2n}{2^n}\le\frac1{2^n}, a convergent geometric series: converges.

Rule of thumb. Look at the dominant behaviour of each term for large nn (here 1n2\frac1{n^2}, 1n3/2\frac1{n^{3/2}}, 1n\frac1n), then find an inequality in the right direction.