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Prove Gauss’ test for convergence of a series of positive terms

Prove Guass’ test for the convergence of a series \sum a_n with a_n > 0 for all n. This test says that if there is an N \geq 1, an s > 1, and an M > 0 such that

    \[ \frac{a_{n+1}}{a_n} = 1 - \frac{A}{n} + \frac{f(n)}{n^s} \qquad \text{for} \qquad n \geq N, \]

where |f(n)| \leq M for all n, then the series \sum a_n converges if A > 1 and diverges if A \leq 1.


Proof. Let \sum a_n be a series of positive terms, and assume there exists an N \geq 1, an s > 1, and an M > 0 such that

    \[ \frac{a_{n+1}}{a_n} = 1 - \frac{A}{n} + \frac{f(n)}{n^s} \]

for all n \geq N, and |f(n)| \leq M for all n. Now we consider the three cases A < 1, A > 1 and A = 1.

Case 1. Assume A < 1. Then we have

    \begin{align*}  \frac{a_{n+1}}{a_n} &= 1 - \frac{A}{n} + \frac{f(n)}{n^s} \\[9pt]  &= 1 - \frac{1+A-1}{n} + \frac{f(n)}{n^2} \\[9pt]  &= 1 - \frac{1}{n} + \frac{1-A}{n} + \frac{f(n)}{n^s}. \end{align*}

Since |f(n)| \leq M for all n \geq N we have

    \[ \lim_{n \to \infty} \frac{f(n)}{n^s} \leq \lim_{n \to \infty} \frac{M}{n^s} = 0. \]

But then since A < 1 we have \frac{1-A}{n} > 0 and so for all sufficiently large n we have

    \[ \frac{1-A}{n} + \frac{f(n)}{n^s} > 0 \]

(since we can make \left| \frac{f(n)}{n^s} \right| arbitrarily small so even if \frac{f(n)}{n^s} < 0 it is smaller in absolute value than \frac{1-A}{n} and so \frac{1-A}{n} + \frac{f(n)}{n^s} must be positive). Since this term is positive we then have

    \[ \frac{a_{n+1}}{a_n} > 1 - \frac{1}{n} \]

for all sufficiently large n. By Raabe’s test (Section 10.16, Exercise #16) we then have \sum a_n diverges.

Case 2. Assume A > 1. Then we have

    \begin{align*}  \frac{a_{n+1}}{a_n} &= 1 - \frac{A}{n} + \frac{f(n)}{n^s} \\[9pt]  &\leq 1 - \frac{A}{n} + \frac{M}{n^s} \\[9pt]  &\leq 1 - \frac{1}{n} + \frac{A-1}{n} + \frac{M}{n^s} \\[9pt]  &\leq 1 - \frac{1}{n} + \frac{A-1+M}{n} &(\frac{M}{n^s} < \frac{M}{n} \text{ since } s>1)\\[9pt]  &\leq 1 - \frac{1}{n} + \frac{r}{n} \end{align*}

for r = A - 1 + M > 0 (since M>0 and A>1). Hence, by Raabe’s test again we have \sum a_n converges.

Case 3. Assume A = 1. From a previous exercise (Section 10.16, Exercise #15) we know that if \{ a_n \} is a sequence of positive terms, \{ b_n \} is a sequence of positive terms such that \sum \frac{1}{b_n} diverges and c_n \leq 0 where

    \[ c_n = b_n - \frac{b_{n+1} a_{n+1}}{a_n} \]

then \sum a_n diverges. So, in this case, let b_{n+1} = n \log n. Then we know from Example 2 on page 398 of Apostol that \sum \frac{1}{b_n} diverges. So, we let

    \begin{align*}  c_n &= b_n - \frac{b_{n+1} a_{n+1}}{a_n} \\[9pt]  &= (n-1)\log(n-1) - n \log n \left( 1 - \frac{1}{n} + \frac{f(n)}{n^s} \right) \\[9pt]  &= (n-1) \log (n-1) - n \log n \left( \frac{n-1}{n} + \frac{f(n)}{n^s} \right) \\[9pt]  &= (n-1) \log (n-1) - \frac{n(n-1)}{n} \log n - \frac{f(n) \log n}{n^{s-1}} \\[9pt]  &= (n-1)(\log (n-1) - \log n) - \frac{f(n) \log n}{n^{s-1}}. \end{align*}

But then, since the logarithm is an increasing function, we know (n-1)(\log (n-1) - \log n) < 0. Furthermore,

    \[ \lim_{n \to \infty} \frac{f(n) \log n}{n^{s-1}} = 0 \]

since |f(n)| \leq M for all sufficiently large n, and \lim_{n \to \infty} \frac{\log n}{n^{s-1}} = 0 by equation (10.11) on page 380 of Apostol. Therefore, for sufficiently large n we have

    \[ (n-1)(\log (n-1) - \log n) - \frac{f(n) \log n}{n^{s-1}} < 0. \]

Hence, by the previous exercise, we have \sum a_n diverges.

So, putting all three cases together we have the requested result, \sum a_n converges if A < 1 and diverges if A \geq 1. \qquad \blacksquare

2 comments

  1. Matt says:

    Regarding case A = 1, showing that \lim_{n \to \infty} \frac{-f(n)\log(n)}{n^{s-1}} = 0 and (n-1)\left( \log(n - 1) - \log(n)\right) < 0 is not sufficient for

        \[\frac{-f(n)\log(n)}{n^{s-1}} + (n-1)\left( \log(n - 1) - \log(n) \right) < 0\]

    In other words showing that a_n < 0 and \lim_{n \to \infty} b_n = 0 does not mean that

        a_n + b_n  0 <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-42e698812184658dafdec5db754e95e1_l3.png" height="14" width="321" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[Hence, before concluding, we must also note that\]" title="Rendered by QuickLaTeX.com"/> 	\lim_{n \to \infty} (n-1)\left( \log(n - 1) - \log(n)\right) = -1 <span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a9aec5f8b810f6025f4f7233ea1c01f8_l3.png" height="43" width="319" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[Which means, for $n > N_1$,\]" title="Rendered by QuickLaTeX.com"/> 	-\frac{3}{2} < (n-1)\left( \log(n - 1) - \log(n)\right)  N_2

    ,

        \[-\frac{1}{2} < \frac{-f(n)\log(n)}{n^{s-1}} < \frac{1}{2}\]

    So, letting N = \max\left\{ N_1, N_2 \right\}

        \[\frac{-f(n)\log(n)}{n^{s-1}} + (n-1)\left( \log(n - 1) - \log(n)\right)< 0\]

    which completes the proof.

    • Matt says:

      I don’t know why it is displayed like that. The Latex is bellow:

      Regarding case $A = 1$, showing that $\lim_{n \to \infty} \frac{-f(n)\log(n)}{n^{s-1}} = 0$ and $(n-1)\left( \log(n – 1) – \log(n)\right) < 0$ is not sufficient for
      $$
      \frac{-f(n)\log(n)}{n^{s-1}} + (n-1)\left( \log(n – 1) – \log(n) \right) < 0
      $$
      In other words showing that $a_n < 0$ and $\lim_{n \to \infty} b_n = 0$ does not mean that $a_n + b_n 0
      $$
      Hence, before concluding, we must also note that
      $$
      \lim_{n \to \infty} (n-1)\left( \log(n – 1) – \log(n)\right) = -1
      $$
      Which means, for $n > N_1$,
      $$
      -\frac{3}{2} < (n-1)\left( \log(n – 1) – \log(n)\right) N_2$,
      $$
      -\frac{1}{2} < \frac{-f(n)\log(n)}{n^{s-1}} < \frac{1}{2}
      $$
      So, letting $N = \max\left\{ N_1, N_2 \right\}$
      $$
      \frac{-f(n)\log(n)}{n^{s-1}} + (n-1)\left( \log(n – 1) – \log(n)\right)< 0
      $$
      which completes the proof.

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