Comparison test for series

For nonnegative sequences with $0\le a_n\le b_n$, convergence of ${\sum }_n{b}_n$ implies convergence of ${\sum }_n{a}_n$, while divergence of ${\sum }_n{a}_n$ implies divergence of ${\sum }_n{b}_n$.

For positive sequences $(a_n),(b_n)$, suppose $a_{n+1}/a_n\le b_{n+1}/b_n$ eventually. Then $a_n/b_n$ is eventually nonincreasing, so $a_n\le C{b}_n$ beyond a fixed index. The comparison test for series proves convergence of $\sum a_n$ whenever $\sum b_n$ converges. Reversing the ratio inequality gives an eventual lower bound $a_n\ge c{b}_n$ and hence divergence when the positive comparison series diverges. The initial finite set of terms does not affect either conclusion.

If $x_n>0$ and $c>0$, clipping every term at $c$ preserves divergence. If infinitely many terms are at least $c$, the clipped series has infinitely many terms equal to $c$ and diverges. Otherwise its tail agrees with the original series. This lemma is useful in proving convergence from minimum-decrement inequalities.

If $f:[1,\infty )\to [0,\infty )$ is decreasing, then ${\sum }_{n\ge 1}f(n)$ and ${\int }_1^{\infty }f(x)dx$ either both converge or both diverge. This follows by bounding the area under the graph between adjacent upper and lower rectangle sums.

For a nonnegative nonincreasing function $f:[1,\infty )\to [0,\infty )$, comparison with unit-width rectangles givesOn $[n,n+1]$, the function is at most $f(n)$; on $[n-1,n]$, it is at least $f(n)$. Summing these inequalities gives the bounds. Applying them to $f(x)=1/x$ gives $\log (N+1)\le H_N\le 1+\log N$ for the harmonic numbers. Unlike a convergence verdict alone, the finite bounds estimate the size of a partial sum.