TITLE: Are telescoping sums related to the fundamental theorem of calculus?
QUESTION [7 upvotes]: I just noticed that
$$\sum_{i=1}^n (a_i - a_{i-1})=a_n-a_0$$
and
$$\int_a^b f'(x)\mathrm{d}x=f(b)-f(a)$$
look really similar.
We can consider $a_i-a_{i-1}$ a discrete analog to the derivative of continuous functions.
Is there anything deep between those equations?

REPLY [3 votes]: Consider
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h$$
So we have
$$\int_a^bf'(x)dx=\lim_{h\to0}\int_a^b\frac{f(x+h)-f(x)}hdx$$
$$=\lim_{h\to0}\frac1h\left(\int_a^bf(x+h)-f(x)dx\right)$$
$$=\lim_{h\to0}\frac1h\left(\int_a^bf(x+h)dx-\int_a^bf(x)dx\right)$$
$$=\lim_{h\to0}\frac1h\left(\int_{a+h}^{b+h}f(x)dx-\int_a^bf(x)dx\right)$$
$$=\lim_{h\to0}\frac1h\left(\int_b^{b+h}f(x)dx-\int_a^{a+h}f(x)dx\right)\tag1$$
$$=f(b)-f(a)\tag2$$
You may graph or look at Riemann sums to see the last step above $(2)$.
Consider your telescoping sum for $(1)$.