Series Proof Workshop I

From Rearrangement Proofs to Convergence Estimates

A rearrangement proof must establish that every index is used and then analyze the reordered partial sums. A related proof skill is useful even when the order of the terms is fixed: separate the oscillation in a sequence from the gradual decay of a weight. The oscillating terms may not themselves tend to zero, and comparison with a positive series may be unavailable, yet their partial sums can remain bounded. When they are multiplied by nonnegative weights that decrease to zero, cancellation can force convergence.

The central result in this workshop is Dirichlet’s test. Its proof illustrates a broader strategy: estimate every finite tail uniformly, then use the Cauchy criterion for series. The key estimate comes from the Finite Summation-by-Parts Identity established earlier in the course. We will use that identity rather than re-prove it.

A Finite Weighted-Tail Estimate

Let $a_n$ be a real sequence, and write $A_0=0$ and $A_n=\sum_{k=1}^{n}a_k$. Suppose the partial sums are bounded: there is an $M\geq0$ such that $|A_n|\leq M$ for all $n\geq0$. Now let $b_n$ be nonnegative, nonincreasing, and tending to zero. The finite sum from $p$ to $q$ can be estimated in terms of the first weight $b_p$, regardless of how many terms the block contains.

Theorem (Finite Weighted-Tail Estimate): Suppose $|A_n|\leq M$ for every $n\geq0$, and $b_n\geq0$ is nonincreasing. For all integers $1\leq p\leq q$, $$ \left|\sum_{n=p}^{q}a_nb_n\right|\leq 2Mb_p. $$

Proof. Apply the Finite Summation-by-Parts Identity to the block from $p$ to $q$, using the partial sums $A_n$. It gives

\sum_{n=p}^{q}a_nb_n = A_qb_q-A_{p-1}b_p+\sum_{n=p}^{q-1}A_n(b_n-b_{n+1}).

Because $b_n$ is nonincreasing, every difference $b_n-b_{n+1}$ in this expression is nonnegative. Taking absolute values and using $|A_n|\leq M$, including $|A_0|=0\leq M$, gives

\left|\sum_{n=p}^{q}a_nb_n\right| \leq Mb_q+Mb_p+M\sum_{n=p}^{q-1}(b_n-b_{n+1}).

The last sum telescopes to $b_p-b_q$; if $p=q$, it is an empty sum equal to zero, and $b_p-b_q=0$ as well. Thus the right-hand side is $Mb_q+Mb_p+M(b_p-b_q)=2Mb_p$. This proves the estimate. $\square$

The estimate is uniform in the endpoint $q$. That is the decisive feature: even a long block of terms cannot produce a large weighted sum if its first weight is small. The next theorem turns this finite estimate into a convergence result.

Dirichlet’s Test

Theorem (Dirichlet’s Test): Suppose the partial sums $A_n=\sum_{k=1}^{n}a_k$ are bounded, and suppose $b_n\geq0$, $b_{n+1}\leq b_n$ for every $n$, and $b_n\to0$. Then the series $\sum_{n=1}^{\infty}a_nb_n$ converges.

Proof. Choose $M\geq0$ such that $|A_n|\leq M$ for every $n\geq0$. If $M=0$, then $A_n=0$ for every $n$, so $a_n=A_n-A_{n-1}=0$ for every $n$, and the series converges. Otherwise, fix $\varepsilon>0$. Since $b_n\to0$, there is an integer $P$ such that $b_p<\varepsilon/(2M)$ whenever $p\geq P$. For any $q\geq p\geq P$, the Finite Weighted-Tail Estimate gives

\left|\sum_{n=p}^{q}a_nb_n\right| \leq 2Mb_p <\varepsilon.

Thus, for every $\varepsilon>0$, all finite sums from $p$ to $q$ have absolute value less than $\varepsilon$ once $q\geq p\geq P$. The Cauchy Criterion for Series now implies that $\sum_{n=1}^{\infty}a_nb_n$ converges. $\square$

Taking the limit as $q\to\infty$ in the finite estimate gives a useful bound on the whole tail. If $S=\sum_{n=1}^{\infty}a_nb_n$, then, for every $p\geq1$,

\left|\sum_{n=p}^{\infty}a_nb_n\right|\leq 2Mb_p.

Indeed, convergence makes the finite sums from $p$ through $q$ tend to the displayed infinite tail, and each finite sum has absolute value at most $2Mb_p$. The estimate is not necessarily sharp, but it is explicit and often enough to control the error after truncation.

Worked Examples: Finding the Bounded Partial Sums

Worked Example: A Sine Series with Reciprocal Square-Root Weights

Consider

\sum_{n=1}^{\infty}\frac{\sin(n\pi/3)}{\sqrt{n}}.

Set $a_n=\sin(n\pi/3)$ and $b_n=1/\sqrt n$. To check the bounded-partial-sum hypothesis, use the identity

2\sin(\theta/2)\sin(k\theta) = \cos((k-\tfrac12)\theta)-\cos((k+\tfrac12)\theta).

Summing this identity for $k=1,\ldots,N$ makes the cosine terms cancel in adjacent pairs, leaving

2\sin(\theta/2)\sum_{k=1}^{N}\sin(k\theta) = \cos(\theta/2)-\cos((N+\tfrac12)\theta).

For $\theta=\pi/3$, the denominator $2\sin(\theta/2)$ equals $1$, and the two cosine values on the right each lie between $-1$ and $1$. Consequently, the partial sums $A_N=\sum_{k=1}^{N}\sin(k\pi/3)$ satisfy $|A_N|\leq2$. The weights $b_n=1/\sqrt n$ are nonnegative, nonincreasing, and tend to zero. Dirichlet’s test proves convergence. The tail estimate also gives, for $p\geq1$,

\left|\sum_{n=p}^{\infty}\frac{\sin(n\pi/3)}{\sqrt n}\right| \leq \frac{4}{\sqrt p}.

Worked Example: Cosines with Slowly Decreasing Weights

Consider the series

\sum_{n=1}^{\infty}\frac{\cos(n\pi/4)}{\log(n+1)}.

Take $a_n=\cos(n\pi/4)$ and $b_n=1/\log(n+1)$. The weights are positive because $n+1>1$, and they decrease: as $n$ increases, $\log(n+1)$ increases. Also, $\log(n+1)\to\infty$, so $b_n\to0$.

For the partial sums of $a_n$, use

2\sin(\theta/2)\cos(k\theta) = \sin((k+\tfrac12)\theta)-\sin((k-\tfrac12)\theta).

Summing from $k=1$ to $N$ gives

2\sin(\theta/2)\sum_{k=1}^{N}\cos(k\theta) = \sin((N+\tfrac12)\theta)-\sin(\theta/2).

With $\theta=\pi/4$, the denominator $2\sin(\pi/8)$ is positive. Since each sine has absolute value at most $1$, we obtain $|A_N|\leq1/\sin(\pi/8)$ for every $N$. All hypotheses of Dirichlet’s test hold, so the series converges. In particular, if $M=1/\sin(\pi/8)$, its tail from $p$ onward has absolute value at most $2M/\log(p+1)$.

Worked Example: A Periodic Sequence with Cancellation

Define $a_n$ by repeating the block $1,-1,-1,1$, and consider

\sum_{n=1}^{\infty}\frac{a_n}{\sqrt n}.

The partial sums of $a_n$ repeat as $1,0,-1,0$: after one block the sum is $1-1-1+1=0$, and within each block the successive partial sums, starting from zero at the block’s beginning, are $1,0,-1,0$. Hence $|A_N|\leq1$ for every $N$. The weights $1/\sqrt n$ are nonnegative, nonincreasing, and tend to zero. Dirichlet’s test proves convergence, and the tail estimate gives

\left|\sum_{n=p}^{\infty}\frac{a_n}{\sqrt n}\right| \leq \frac{2}{\sqrt p}.

The block calculation is used only to bound the partial sums; the weights are not constant within a block. Therefore one should not replace each weighted block by an unweighted block sum and declare it zero.

What the Test Does—and Does Not—Say

Dirichlet’s test is useful when the factors have different jobs. The sequence $a_n$ supplies cancellation through bounded partial sums, while $b_n$ supplies decay through monotonicity and convergence to zero. Neither factor needs to define a convergent series on its own. In particular, bounded partial sums of $a_n$ do not mean that $\sum a_n$ converges; the test controls the weighted series.

The hypotheses should be checked as stated. It is not enough that $b_n\to0$ if the weights oscillate rather than decrease: the finite estimate depends on $b_n-b_{n+1}\geq0$. Likewise, decreasing weights alone do not guarantee convergence if the partial sums of $a_n$ are unbounded. The proof needs both controls.

Finally, Dirichlet’s test proves ordinary convergence, not absolute convergence. It bounds sums with their signs and cancellation intact. To establish absolute convergence, one must separately analyze $\sum |a_nb_n|$, using an appropriate comparison or another convergence test. Keeping these conclusions distinct is essential whenever oscillation is doing the work.

Takeaway: If the partial sums of $a_n$ are bounded and $b_n$ is nonnegative, nonincreasing, and tends to zero, then $\sum a_nb_n$ converges. The finite-tail bound $2Mb_p$ makes the mechanism quantitative.

Check Your Understanding

Use the estimates and proof strategy in this workshop to answer the following questions.

In the Finite Weighted-Tail Estimate, where does the nonincreasing property of $b_n$ enter the proof?
Why does the bound $2Mb_p$ imply the Cauchy condition when $b_p\to0$?
For $a_n=\cos(n\theta)$, what condition on $\theta$ is needed to use the displayed cosine-sum identity as a bounded-partial-sum estimate?
Why does Dirichlet’s test not, by itself, establish absolute convergence?
For the repeating block $1,-1,-1,1$, what are the possible values of its partial sums?

What You'll Learn