Show: $\sum_{n=1}^{k}\sum_{m=n}^{k}\int_m^{m+1}f(x)dx=\sum_{m=1}^{k}m \int_m^{m+1}f(x)dx$

Question

What I understand is that $\sum_{n=1}^{k}\sum_{m=n}^{k}\int_m^{m+1}f(x)dx$ means summing $\int_m^{m+1}f(x)dx$ from $1$ to $k$ twice, since we see that $\sum_{n=1}^{k}=\sum_{m=1}^{k}$ since, $m=n$, but then how we get the identity given above? I am confused!

Please show elaborately that-

$$\sum_{n=1}^{k}\sum_{m=n}^{k}\int_m^{m+1}f(x)dx$$ $$=\sum_{m=1}^{k}m \int_m^{m+1}f(x)dx$$

Thanks.

Answer 1

It is just a matter of definition. $$ \sum_{n=1}^{k} \sum_{m=n}^{k} \int_m^{m+1} f(x)dx $$ Expand along the first sum: $$ \left( \sum_{m=1}^{k} \int_m^{m+1} f(x)dx \right) + \left( \sum_{m=2}^{k} \int_m^{m+1} f(x)dx \right) + \cdots + \left( \sum_{m=k}^{k} \int_m^{m+1} f(x)dx \right) $$ Write each term in parentheses as a new line, and expand the sum: $$ \begin{array}{ccccccc} \int_1^{2} f(x)dx & + & \int_2^{3} f(x)dx & + & \cdots & + & \int_k^{k+1} f(x)dx \\ & + & \int_2^{3} f(x)dx & + & \cdots & + & \int_k^{k+1} f(x)dx \\ &&&& \ddots && \vdots \\ &&&&& + & \int_k^{k+1} f(x)dx \end{array} $$ Now count the number of each integral: $$ 1 \cdot \left( \int_1^{2} f(x)dx \right) + 2 \cdot \left( \int_2^{3} f(x)dx \right) + \cdots + k \cdot \left( \int_k^{k+1} f(x)dx \right) $$ Recombine using sigma notation: $$ \sum_{m=1}^{k} m \int_m^{m+1} f(x)dx $$

Answer 2

In general for a sequence $\{a_m\}$:

$$\sum_{n=1}^{k}\sum_{m=n}^{k} a_m =\sum_{m=1}^{k}m a_m.$$

That can be proven by induction on $k$.

Replace $a_m$ with $\int_m^{m+1}f(x)$... and you're done.

Answer 3

Another proof without induction

How many points is there in each horizontal line?

Answer 4

With $a_m = \int_m^{m+1} f(x)\,dx$, the sum can be written as

$$\sum_{n=1}^k\sum_{m=1}^k \mathbf{1}_{m\geqslant n}a_m = \sum_{m=1}^k \left(\sum_{n=1}^k \mathbf{1}_{m\geqslant n}\right) a_m = \sum_{m=1}^k m a_m$$

Answer 5

We can write the index region conveniently to better see what's going on.

We obtain

\begin{align*} \sum_{n=1}^k\sum_{m=n}^k\int_{m}^{m+1}f(x)\,dx&=\sum_{\color{blue}{1\leq n\leq m\leq k}}\int_{m}^{m+1}f(x)\,dx\tag{1}\\ &=\sum_{m=1}^k\sum_{n=1}^m\int_{m}^{m+1}f(x)\,dx\tag{2}\\ &=\sum_{m=1}^k\int_{m}^{m+1}f(x)\,dx\sum_{n=1}^m1\tag{3}\\ &=\sum_{m=1}^km\int_{m}^{m+1}f(x)\,dx \end{align*}

Comment:

• (1): The inequality chain shows the relationship of the indices $n$ and $m$,

• (2): enabling us to exchange the sums easily.

• (3): The integral does not depend on the index $n$ and can be factored out.