Determine the infinite sum of b_k = c_k - c_(k-1)

[lizarton]

Suppose that $b_k = c_k - c_{k-1}$, where $\langle c \rangle$ is a sequence such that $c_0=1$ and $\lim_{k → ∞} c_k = 0$. Use the definition of series to determine $\sum^{∞}_{k=1} b_k.$

I've done a little analysis, so I think that $c_k$ is decreasing, since the first term is greater than the limit of the sequence $\Rightarrow b_k$ $\leq 0$ and decreasing (I think).

Also, by taking limit of both sides I obtain:
$\lim_{k → ∞} b_k = \lim_{k → ∞} c_k - c_{k-1} = 0$.
(I understand this does not conclude that the series converges)

From the looks of the problem, it smells like Cauchy convergence criterion, but I'm not sure where to start with such an abstract description of the sequences.

The hint says to compute the partial sums exactly, but since no explicit formula is given for $\langle c \rangle$, I'm not so sure how to go about this.

Here is the definition we use for an infinite series:

Let $\langle a \rangle$ be a sequence of real numbers. The formal expression $\sum^{∞}_{k=1} a_k$ is an infinite series. The number $s_n = \sum^{n}_{k=1} a_k$ is the nth partial sum of the series. The infinite series $\sum^{∞}_{k=1} a_k$ converges if $\lim_{n → ∞} s_n$ exists; otherwise the series diverges. When $\sum^{∞}_{k=1} a_k$ converges, we write $L = \lim s_n=\sum^{∞}_{k=1} a_k$ and say that the sum of the series equals $L$.

Any help would be appreciated. Thanks so much!

*EDIT* Let me know if I'm onto something here:
Let $s_n = \sum^{n}_{k=1} b_k$ and $S_n = \sum^{n}_{k=0} c_k = 1 + \sum^{n}_{k=1} c_k.$

Then $S_{n-1} = \sum^{n-1}_{k=0} c_{k-1} = 1 + \sum^{n-1}_{k=1} c_k.$

Then $s_n = S_n - S_{n-1} = 1 + \sum^{n}_{k=1} c_k - 1 - \sum^{n-1}_{k=1} c_k = c_n.$

$\sum^{∞}_{k=1} b_k = \lim_{n→∞} (S_n - S_{n-1}) = \lim_{n→∞} c_n = 0$

 

[micromass]

First of all, the $c_n$ does not need to be decreasing.

Second, the hint is a good one:

$$b_0=c_0$$

$$b_1+b_0=(c_1-c_0)+c_0=c_1$$

$$b_2+b_1+b_0=(c_2-c_1)+(c_1-c_0)+c_0=c_2$$

Generalize.

EDIT: it seems you found it already...