How do I prove convergence of the recursive sequence $c_n = c_{n-1} + \frac{0.01}{n}$?

Question

I have this sequence: $c_n = c_{n-1} + \frac{0.01}{n}, \ \ c_1 = 0.01$

How do I prove the convergence of this, and what is the limit?

Context

I was trying to solve the problem of a snail crawling on an elastic band being stretched. I wrote a small computer program to simulate this, and since a large number of iterations did not produce a large change, I wrongly guessed this would converge (also I was biased due to a discussion on the problem). I considered this a sequence, which was one of the reasons I got stuck in trying to prove this. As pointed out, I should have written it as a sum.

Answer 1

No limit exists since this is not convergent. Because $1/n > \ln(1+1/n)$ (because $e^x>1+x$ when $x>0$.) Then $\sum_{n=1}^k 1/n > \sum_{n=1}^k \ln(n+1)-\ln(n) = \ln(k+1)\to\infty$

Answer 2

Let us consider the more general case $$c_n = c_{n-1} + \frac{a}{n}, \ \ c_1 = b$$ the first terms of the sequence will be $b$, $b+\frac{a}{2}$, $b+\frac{5 a}{6}$, $b+\frac{13 a}{12}$ and more genrally $$c_n=b+a (H_n-1)$$ which generalizes what David Mitra commented.

Then, just as its cousin, it does not converge.

Answer 3

As Claude Leibovici said, the sequence does not converge. That is because the harmonic series $H_n=\sum\limits_{k=1}^n\frac{1}{k}$ diverges. We can see that because if we group the terms of the sum we see it is greater than an infinite sum of halves, which obviously diverges. The first term is 1, and we leave it stand-alone. The second one is a half, and we leave it like that. The third term is $\frac{1}{3}>\frac{1}{4}$, which summed with the following quarter gives a half. The following four are all not less than an octave, and four octaves is a half. The following eight are all not less than a sixteenth, and $\frac{8}{16}=\frac{1}{2}$. And so on.

Answer 4

Another slightly different approach.

We know that $ c_n - c_{n-1 } = \dfrac{0.01}{n} \implies c_n - c_{n- 1 } = \dfrac{0.01}{n} $

Let $ m \gt n \ge N $. Then, $$ |c_m - c_n| = |(c_m - c_{m- 1}) + (c_{m - 1} - c_{m- 2} ) + ... + (c_{n+ 1} - c_n)| $$

$$ = 0.01| \frac{1}{m} + \frac{1}{m -1} + ... + \frac{1}{n + 1} | $$

Especially, when $m = 2n$; $$ = 0.01 | \frac{1}{2n} + ... \frac{1}{n + 1} | \ge \; \; 0.01 \times n \times \frac{1}{2n} = 0.005 $$

Hence the given sequence is not Cauchy and hence cannot converge.