Q. 79 - Calculus

Question

For any positive integer k, the following equation holds:

$1+2+3+···+k=\frac{k(k + 1)}{2}.$ Use this fact to prove that for all k > 100, the value of the sum of the first k integers is greater than 5000. What does this have to do with the limit of a sequence of sums as $k\to \infty ?$

Step-by-Step Solution

Answer

The given statement is proved. The limit of a sequence of sums as $k\to \infty$ will also be $\infty .$

Step 1. Given Information.

The given equation is $1+2+3+···+k=\frac{k(k + 1)}{2}.$

Step 2. Prove.

Let the function is $f\left(k\right)=\frac{k(k+1)}{2}.$

Take the limit of the above function as $k\to 101.$

$$\begin{gathered} \underset{k\to 101}{\lim }f\left(k\right)=\underset{k\to 101}{\lim }\frac{k(k+1)}{2} \\ \underset{k\to 101}{\lim }f\left(k\right)=\frac{101(101+1)}{2} \\ \underset{k\to 101}{\lim }f\left(k\right)=\frac{101\left(102\right)}{2} \\ \underset{k\to 101}{\lim }f\left(k\right)=101\left(51\right) \\ \underset{k\to 101}{\lim }f\left(k\right)=5151 \end{gathered}$$

Similarly for the other larger values of k, the value of the function will be even larger.

Therefore, the limit of a sequence of sums as $k\to \infty$ will also be $\infty .$