1

Given only that the series $\sum a_n$ converges, either prove that the series $\sum b_n$ converges or give a counterexample, when we define $b_n$ by,

i ) $\frac{a_n}{n}$

ii) $a_n \sin(n ) $

iii) $n^{\frac{1}{n}} a_n$

Is there any general approach to such questions?

Michael Rozenberg
  • 203,855
Limzy
  • 168

3 Answers3

1

Counterexample (ii):

The series

$$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty\frac{\sin n}{n},$$

converges by the Dirichlet test. Using $\cos 2n = 1 - 2 \sin^2 n$ we have

$$\sum_{n=1}^m a_n \sin n = \sum_{n=1}^m\frac{\sin^2 n}{n} = \sum_{n=1}^m\frac{1}{2n}- \sum_{n=1}^m\frac{\cos 2n}{2n}$$

The first sum on the RHS diverges (harmonic series) and the second sum converges by the Dirichlet test. Hence, the sum on the LHS diverges.

I'll leave it to you to attempt part (i) and (iii) where a standard series test applies. Hint -- the sequence $1/n$ is monotone decreasing and $n^{1/n}$ is eventually monotone decreasing. Case (ii) stands apart because $\sin n$ is not monotone.

Supplement: Deriving $\sum_{n=1}^N \sin n = \frac{\sin (N/2) \sin( (N+1)/2)}{\sin (1/2)}$

We will use the identities $\sin a = - \cos (\pi/2 + a)$ and $\sin(a+b) - \sin(a-b) = 2 \sin b \cos a. $

We have

$$\begin{align}2 \sin(1/2) \sum_{n=1}^N \sin n &= -\sum_{n=1}^N [2 \sin(1/2) \cos(\pi/2 +n)] \\ &= -\sum_{n=1}^N [ \sin(\pi/2 + n + 1/2) - \sin(\pi/2 +n - 1/2)] \end{align}.$$

Since the series on the RHS is telescoping, we get

$$\begin{align}2 \sin(1/2) \sum_{n=1}^N \sin n &= - [\sin(\pi/2 + N + 1/2) - \sin(\pi/2 + 1/2)] \\ &= -[ \sin( N/2 + 1/2 + \pi/2 + N/2 ) - \sin(N/2 + 1/2 + \pi/2 - N/2)] \end{align}$$

Using $\sin(a+b) - \sin(a-b) = 2 \sin b \cos a $ with $a = N/2 + 1/2 + \pi/2$ and $b = N/2$,

$$\begin{align}2 \sin(1/2) \sum_{n=1}^N \sin n &= - 2\sin(N/2) \cos(\pi/2 + (N+1)/2) \\ &= 2\sin(N/2) \sin((N+1)/2) \end{align}$$

Hence,

$$\sum_{n=1}^N \sin n = \frac{2\sin(N/2) \sin((N+1)/2)}{\sin(1/2)}$$

RRL
  • 92,835
  • 7
  • 70
  • 142
  • Hi, I am unfamiliar with the Dirichlet test, could you elaborate more please? – Limzy Mar 17 '17 at 03:17
  • Dirichlet's test states that if $b_n$ decreases monotonically to $0$ and the partial sums $\sum_{n=1}^N a_n$ are bounded (not necessarily convergent), then $\sum a_n b_n$ is convergent. After the alternating series test (which is a special case) it is one of the first tests to try in checking conditional convergence. A related test is Abel's test which is what you should research to finish (i) and (iii). That is the "general approach" to consider for such problems when we are given only that $\sum a_n$ converges conditionally. – RRL Mar 17 '17 at 03:22
  • So in the case of ii) we take $b_n$ to be $\frac{1}{n}$ and $a_n$ to be $sin (n)$ ? – Limzy Mar 17 '17 at 03:24
  • Yes and $\sum_{n=1}^N \sin n = \sin(N/2) \sin((N+1)/2) / \sin(1/2)$ which is bounded. – RRL Mar 17 '17 at 03:26
  • That just shows that $\sum a_n $ where $a_n = \sin n / n$ converges. The rest of my answer explains why $\sum a_n \sin n$ diverges to provide a counterexample. – RRL Mar 17 '17 at 03:29
  • I have been referring to http://math.stackexchange.com/a/1206665/376934 for an explanation on how to obtain the partial sum of $sin (n)$. However, the step to get from $Im(\frac{e^i - e^{(N+1)i}}{1-e^i})$ to $ sin(N/2)sin((N+1)/2)/sin(1/2)$ seems very tedious to compute by expansion. Are there any tricks we can use? – Limzy Mar 17 '17 at 04:39
  • This is a bit of a diversion for this problem, but it's great you are interested. I derived the sum, as an addition to my answer, using only "basic" trigonometry and many "tricks." I like this better than using the complex exponential but it is not easy either. Finally you should be sure to understand why (i) and (iii) converge -- Abel's test holds because $\sum a_n$ converges and in each case $b_n$ is bounded and monotone (eventually monotone for $n^{1/n})$. – RRL Mar 17 '17 at 05:19
  • Let us continue this discussion in chat. – Limzy Mar 17 '17 at 05:55
  • OK -- are you in chat now? – RRL Mar 17 '17 at 06:02
0

i). $b_n = a_n / n$

$n$ is always positive in this summation, so $b_n$ and $a_n$ have the same sign.

If $\sum_{n=1}^{\infty} a_n$ converges absolutely, then $0 \leq |a_n|/n \leq |a_n| \ \ \ \forall n\geq 1$ so $ \sum b_n$ also converges absolutely by the squeeze or sandwich theorem.

If $\sum_{n=1}^{infty} a_n$ converges conditionally, then (a) $b_n$ also has strictly alternating signs, (b) $\lim_{n \rightarrow \infty} a_n = 0$ implies $\lim_{n \rightarrow \infty} a_n/n = b_n = 0$, (c) $ \forall n, |a_{n+1}| < |a_n|$ implies $\forall n, |a_{n+1}|/n < |a_n|/n$ and $b_{n+1}=|a_{n+1}|/(n+1) < |a_{n+1}|/n < |a_n|/n = b_n$ so $\sum b_n$ also converges conditionally.

(ii) $b_n = a_n \sin (n)$

If $a_n \geq 0 \ \ \ \forall n \geq 1,$ OR if $\sum a_n $ converges absolutely, then since $-1 \leq \sin (n) \leq 1 \implies 0 \leq |\sin (n) | \leq 1$

Thus $0 \leq |b_n| = |\sin (n)| |a_n| \leq |a_n| $ so $\sum b_n$ converges absolutely by the squeeze or sandwich theorem..

BUT if $\sum a_n$ converges only conditionally, in this case we canot be sure about $ \sum b_n$

(iii) $b_n = n^{1/n} a_n$

$ n = e^{\ln n} $ so $ n^{1/n} = e^{\ln n \cdot 1/n} = e^{\frac{\ln n}{n}}$

$ \ln n < n \ \ \ \ \forall n > 0$ and by l'Hopital or other methods $ \lim_{n \rightarrow \infty} (\ln n) / n = 0$, so $ \lim_{n \rightarrow \infty} e^{\frac{\ln n}{ n}} = e^0 = 1$

Thus $\lim_{n \rightarrow \infty} b_n = a_n$ and therefore $\sum b_n$ converges by the limit comparison test.

Yes there is a general method. Use the squeeze or sandwich theorem and comparison test where possible and use limits and the limit comparison tests otherwise. It helps to know properties of trig functions and exponents and other common functions in the limit. Be very careful with signs.

victoria
  • 1,429
  • Whoever's voting down, you could say why. I do see a lot of typos here which I am now going to fix; didn't have time before. – victoria Mar 20 '17 at 01:27
-1

For (iii) consider the sequence to be $n^{-n}$

Nosrati
  • 30,522
DrMathy
  • 1