ANOTHER ELEMENTARY PROOF OF THE CONVERGENCE–DIVERGENCE OF p–SERIES

Rasul A. Khan

Recently Khan [2] gave a simple proof of the convergence–divergence of the p–series ∑_{n=1}^{∞} 1/n^{p}. The divergence of this series for p ≤ 1 was shown by contradiction while the convergence of the series for p > 1 was established by the boundedness of the monotonic partial sums [2]. Here, we give a more direct and very elementary proof of the same by using only the sum of a geometric series. Moreover, a telescoping method is used to find sums of some interesting series.

We use the following simple fact:

∑_{n=1}^{∞} ar^{n−1} = \frac{a}{1 − r} , |r| < 1.  (1)

We consider integers j from 2^{m} to 2^{m+1} − 1 (m = 0, 1, 2, …), and note that the number of terms are 2^{m+1} − 1 − (2^{m} − 1) = 2^{m+1} − 2^{m} = 2^{m}. Then, for any p we write the p–series as

∑_{n=1}^{∞} \frac{1}{n^{p}} = ∑_{m=0}^{∞} ∑_{j=2^{m}}^{2^{m+1}−1} \frac{1}{j^{p}} .  (2)

If p > 1, it then follows from (1) and (2) that

∑_{n=1}^{∞} \frac{1}{n^{p}} ≤ ∑_{m=0}^{∞} \frac{2^{m}}{2^{mp}} = ∑_{m=0}^{∞} \frac{1}{2^{m(p−1)}} = \frac{2^{p−1}}{2^{p−1} − 1},

and the series converges. To show the divergence for p ≤ 1, we consider p = 1 first. It is clear from (2) that

∑_{n=1}^{∞} \frac{1}{n} = ∑_{m=0}^{∞} ∑_{j=2^{m}}^{2^{m+1}−1} \frac{1}{j} ≥ ∑_{m=0}^{∞} \frac{2^{m}}{2^{m+1}} = ∞.

If p ≤ 1, then ∑_{n=1}^{∞} 1/n^{p} ≥ ∑_{n=1}^{∞} 1/n = ∞, and the series diverges. Hence, the convergence–divergence of the p–series has been established for every value of p.

The preceding observation can be further used to determine the convergence/divergence of some other series ∑_{n=k}^{∞} f(n) (k ≥ 1), where f(x) is a decreasing positive function. For example, we have

∑_{n=2}^{∞} \frac{1}{n(ln n)^{p}} = ∑_{m=1}^{∞} ∑_{j=2^{m}}^{2^{m+1}−1} \frac{1}{j(ln j)^{p}} ,

and

∑_{n=2}^{∞} \frac{1}{n(ln n)^{p}} ≤ ∑_{m=1}^{∞} \frac{2^{m}}{2^{m}(m ln 2)^{p}} = ∑_{m=1}^{∞} \frac{1}{c m^{p}} , c = (ln 2)^{p},

and the series converges if p > 1 by the p–series. Moreover,

∑_{n=2}^{∞} \frac{1}{n ln n} ≥ ∑_{m=1}^{∞} \frac{2^{m}}{2^{m+1} ln 2^{m+1}} = \frac{1}{2} ∑_{m=1}^{∞} \frac{1}{(m + 1) ln 2} = +∞,

and thus, ∑_{n=2}^{∞} 1/(n(ln n)^{p}) diverges for p ≤ 1. Perhaps, even more interesting is the series ∑_{n=2}^{∞} 1/((ln n)^{p}) for any p > 0. Clearly,

∑_{n=2}^{∞} \frac{1}{(ln n)^{p}} = ∑_{m=1}^{∞} ∑_{j=2^{m}}^{2^{m+1}−1} \frac{1}{(ln j)^{p}} ≥ ∑_{m=1}^{∞} \frac{2^{m}}{(ln 2^{m+1})^{p}} = ∑_{m=1}^{∞} \frac{2^{m}}{(m + 1)^{p}(ln 2)^{p}} .

Since lim_{m→∞} 2^{m}/(m^{p}) = ∞ (m − p ln m = O(m) and 2^{m−p ln m} → ∞), the series ∑_{n=2}^{∞} 1/((ln n)^{p}) diverges for every fixed p.

It is interesting to note that the telescoping method can be used to find the sum of the series ∑_{n=1}^{∞} n/(2^{n}) and ∑_{n=1}^{∞} n^{2}/(2^{n}), etc. However, as suggested by a colleague we consider a more general case, namely, ∑_{n=1}^{∞} n^{k}r^{n}, |r| < 1. First we observe that the series converges absolutely. To see this let r ≠ 0 and note that |r|^{n} = exp(−αn), where α = − ln |r| > 0. Since exp(αn) > (α^{k+2}n^{k+2})/((k + 2)!), hence, n^{k}|r|^{n} = n^{k}exp(−αn) < λ_{k}/(n^{2}), for all n ≥ 1, where λ_{k} = ((k + 2)!)/(α^{k+2}). Thus, ∑_{n=1}^{∞} n^{k}|r|^{n} ≤ λ_{k} ∑_{n=1}^{∞} 1/(n^{2}) < ∞ by the p–series. Therefore, letting a_{n} = n^{k}r^{n}, lim_{n→∞} n^{k}r^{n} = lim_{n→∞} a_{n} = 0. It then follows that ∑_{n=1}^{∞}(a_{n} − a_{n+1}) converges, and ∑_{n=1}^{∞}(a_{n} − a_{n+1}) = a_{1}. Now set S(k) = ∑_{n=1}^{∞} a_{n} = ∑_{n=1}^{∞} n^{k}r^{n}. Clearly, S(0) = r/(1 − r), and applying the binomial theorem we have

a_{n} − a_{n+1} = n^{k}r^{n} − (n + 1)^{k}r^{n+1} = (1 − r)n^{k}r^{n} − ∑_{j=0}^{k−1} \binom{k}{j}n^{j}r^{n+1}.

Now summing over n and using the telescoping sum we obtain

\begin{aligned}r & = a_{1} = ∑_{n=1}^{∞}(a_{n} − a_{n+1}) = ∑_{n=1}^{∞}(1 − r)n^{k}r^{n} − ∑_{n=1}^{∞} ∑_{j=0}^{k−1} r\binom{k}{j}n^{j}r^{n} \\ & = (1 − r) ∑_{n=1}^{∞} n^{k}r^{n} − r ∑_{j=0}^{k−1} \binom{k}{j} ∑_{n=1}^{∞} n^{j}r^{n} = (1 − r)S(k) − r ∑_{j=0}^{k−1} \binom{k}{j}S(j). \end{aligned}

Hence, the preceding equation gives

S(k) = \frac{r}{1 − r} (1 + ∑_{j=0}^{k−1} \binom{k}{j}S(j)) = S(0)(1 + ∑_{j=0}^{k−1} \binom{k}{j}S(j)).

In particular, for r = 1/2, using the recurrence we obtain

\begin{aligned}S(0) & = ∑_{n=1}^{∞} \frac{1}{2^{n}} = 1, S(1) = ∑_{n=1}^{∞} \frac{n}{2^{n}} = 2, \\ S(2) & = ∑_{n=1}^{∞} \frac{n^{2}}{2^{n}} = 6, S(3) = ∑_{n=1}^{∞} \frac{n^{3}}{2^{n}} = 26, etc. \end{aligned}

It is tempting to find a closed formula for S(k), but unfortunately it appears that such a formula is intractable and cannot be obtained even for the special case r = 1/2.

\underline{References}

1. T. Cohen and W. J. Knight, “Convergence and Divergence of ∑_{n=1}^{∞} 1/n^{p}, Mathematics Magazine, 52 (1979), 178.

2. R. A. Khan, “Convergence–Divergence of p–Series,” College Mathematics Journal, 32 (2001), 206–207.

Rasul A. Khan Department of Mathematics Cleveland State University Cleveland, OH 44115 email: khan@math.csuohio.edu