Question

For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[infinity]Sn to obtain the value of the series or state that the series diverges.
1) ∑[infinity]k = 1(1/k + 1 - 1/k + 2).
2) ∑[infinity]k = 1(1/(k + 6)(k + 7).

Answer 1

Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

$1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

Solve point 1 that is $\sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}$:

when,

$k= 1 \to s_1 = \frac{1}{1+1} - \frac{1}{1+2}$

                  $= \frac{1}{2} - \frac{1}{3}$

$k= 2 \to s_2 = \frac{1}{2+1} - \frac{1}{2+2}$

                  $= \frac{1}{3} - \frac{1}{4}$

$k= 3 \to s_3 = \frac{1}{3+1} - \frac{1}{3+2}$

                  $= \frac{1}{4} - \frac{1}{5}$

$k= n^ \to s_n = \frac{1}{n+1} - \frac{1}{n+2}$

Calculate the sum $(S=s_1+s_2+s_3+......+s_n)$

$S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}$

   $=\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}$

When $s_n \ \ dt_{n \to 0}$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}$

$\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}$

In point 2: $\sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

when,

$k= 1 \to s_1 = \frac{1}{(1+6)(1+7)}$

                  $= \frac{1}{7 \times 8}\\\\= \frac{1}{56}$

$k= 2 \to s_1 = \frac{1}{(2+6)(2+7)}$

                  $= \frac{1}{8 \times 9}\\\\= \frac{1}{72}$

$k= 3 \to s_1 = \frac{1}{(3+6)(3+7)}$

                  $= \frac{1}{9 \times 10} \\\\ = \frac{1}{90}$

$k= n^ \to s_n = \frac{1}{(n+6)(n+7)}$

calculate the sum:$S= s_1+s_2+s_3+s_n$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}$

when $s_n \ \ dt_{n \to 0}$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066$

$\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}$