Let $\lim_{n \to \infty} a_n=A$. Show that if $b_n=a_{n+1}-a_n$, then $S=\sum_{n=1}^\infty b_n$ is convergent. 
Let $(a_n)_{n=1}^\infty$ be a sequence with $\lim_{n \to \infty} a_n=A$. Show that if $b_n=a_{n+1}-a_n$, then the infinite series $$S=\sum_{n=1}^\infty b_n$$ is convergent. What is $S$ equal to?


Suppose $b_n=a_{n+1}-a_n$. Then
$$\lim_{n\to \infty} b_n = \lim_{n\to \infty} (a_{n+1}-a_n)=\lim_{n\to \infty} a_{n+1}-\lim_{n\to \infty} a_n=A-A=0$$
Since $b_n \to 0$, we can consider checking if $S$ is convergent. 
Since $b_n=a_{n+1}-a_n$, then 
\begin{align*}
 S&=\sum_{n=1}^\infty (a_{n+1}-a_n)\\
 &=({a_2}-a_1)+({a_3}-{a_2})+({a_4}-{a_3})+({a_5}-{a_4})+\cdots\\
 &=-a_1
\end{align*}
So, $S=-a_1$. Thus, $S$ is convergent.
Is this correct?
 A: It is not equal to $-a_1$. Did you not find weird that $A$ didn't appear in your result? Just note that  $$\sum_{n=1}^N b_n = a_{N+1}-a_1\to A-a_1.$$
A: Let $(s_n)$ be the sequence of partial sums for $\sum b_k$. Then for every $n \geq 1$, \begin{align*}s_n = \sum_{k=1}^{n} b_k &= \sum_{k=1}^{n} (a_{k+1}-a_k) \\&= (a_2-a_1)+(a_3-a_2)+\cdots + (a_{n+1}-a_n)\\[1ex]&= a_{n+1}-a_1.\end{align*} Since $\lim a_{n+1}=\lim a_n = A$, $$\sum_{k=1}^{\infty} b_k = \lim s_n = \lim {}(a_{n+1}-a_1) = \lim a_{n+1}-\lim a_1 = A - a_1.$$ So $\sum b_k$ indeed converges; its sum is $A - a_1$.
A: A series is convergent if and only if its sequence of partial sums is convergent.
Let $(B_N)$ be the sequence of partial sums and consider, for sake of an example, the partial sum $B_4$:
\begin{align*}
B_4 &=\sum_{n=1}^4 b_n\\
&=\sum_{n=1}^4 a_{n+1}-a_n\\
&=a_2-a_1+a_3-a_2+a_4-a_3+a_5-a_4\\
&=-a_1+a_5\\
&=a_5-a_1.
\end{align*}
Now, if we consider any partial sum $B_N$, we can see that $B_N=a_{N+1}-a_1$.
So, 
$$\lim_{N\to \infty} B_N=\lim_{N\to \infty} a_{N+1}-a_1=A-a_1.$$
So, the sequence of partial sums $(B_N)$ converges to $A-a_1$ and thus 
$$S=\sum_{n=1}^\infty b_n=\lim_{N\to \infty} \sum_{n=1}^N b_n=\lim_{N\to \infty}B_N=A-a_1.$$
A: For $1 \leq k < \infty$, let
$$S_k = \sum_{n=1}^{k}b_n\text{.}$$
Regardless of the value of $k$, we have
$$S_k = b_1+b_2 + \cdots + b_k = a_2 - a_1+a_3-a_2+\cdots+a_{k+1}-a_k=a_{k+1}-a_1\text{.}$$
The sum you desire, $S$, is equal to $S_{\infty} = \lim\limits_{k \to \infty}S_k$. Observe that $a_1$ is constant with respect to $k$. Given that $\lim\limits_{k \to \infty}a_{k+1} = \lim\limits_{k \to \infty}a_{k}=A$, we obtain
$$S = S_{\infty} = \lim\limits_{k \to \infty}(a_{k+1}-a_1)=\left(\lim\limits_{k \to \infty}a_{k+1}\right)-a_1=A-a_1\text{.}$$
