# Riemann sum: sum_riemann

### 5.20.4 Riemann sum: sum_riemann

Given a function f on [0,1], the Riemann sum corresponding to dividing the interval into n equal parts and using the right endpoints is

n
∑
k=1

f(

x
n

)

1
n

.

The sum_riemann command determines if a sum is such a Riemann sum, and if it is, evaluates the integral.

- sum_riemann takes two arguments:
  - expr, an expression depending on two variables.
  - [n,k], the list of those two variables.
- sum_riemann(expr,[n,k]) returns

  lim
  n→∞

  n
  ∑
  k=1

  expr

  (which, viewing the sum as a Riemann sum of a continuous function on [0,1], is the definite integral) or returns "it is probably not a Riemann sum" when the no result is found.

Exercises.

1. Suppose Sn=∑k=1n k2/n3.
   Compute limn → +∞ Sn.
   Input:
   sum_riemann(k^2/n^3,[n,k])
   Output:

   1
   3

2. Suppose Sn=∑k=1n k3/n4.
   Compute limn → +∞ Sn.
   Input:
   sum_riemann(k^3/n^4,[n,k])
   Output:

   1
   4

3. Compute limn → +∞(1/n+1+ 1/n+2+…+1/n+n).
   Input:
   sum_riemann(1/(n+k),[n,k])
   Output:

   ln | ⎛ ⎝ | 2 | ⎞ ⎠

4. Suppose Sn=∑k=1n 32n3/16n4−k4.
   Compute limn → +∞ Sn.
   Input:
   sum_riemann(32*n^3/(16*n^4-k^4),[n,k])
   Output:

   2 arctan

   ⎛
   ⎜
   ⎜
   ⎝

   1
   2

   ⎞
   ⎟
   ⎟
   ⎠

   +ln

   ⎛
   ⎝

   3

   ⎞
   ⎠