How do you find the 4-th partial sum of the infinite series #sum_(n=1)^oo(1/sqrt(n)-1/sqrt(n+1))# ?

Question

Samantha Adkison · Accepted Answer

undefined To find the fourth partial sum of the given infinite series $ \sum_{n=1}^{\infty} \left(\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}\right) $, you need to evaluate the expression for the first four terms of the series and then add them together.

The general term of the series is $ a_n = \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} $.

Now, compute $ a_1 $, $ a_2 $, $ a_3 $, and $ a_4 $, and then sum them up to find the fourth partial sum.

$$
a_1 = \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}}
$$

$$
a_2 = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}
$$

$$
a_3 = \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}}
$$

$$
a_4 = \frac{1}{\sqrt{4}} - \frac{1}{\sqrt{5}}
$$

Add these values together to find the fourth partial sum.

Scarlett Allison · Answer

#S_4=(1/sqrt{1}-1/sqrt{2})+(1/sqrt{2}-1/sqrt{3})+(1/sqrt{3}-1/sqrt{4})+(1/sqrt{4}-1/sqrt{5})#

by regrouping,

#=1+(-1/sqrt{2}+1/sqrt{2})+(-1/sqrt{3}+1/sqrt{3})+(-1/sqrt{4}+1/sqrt{4})-1/sqrt{5}#

by cancelling intermediate terms,

#=1-1/sqrt{5}#

Charles Arocha · Answer

undefined To find the fourth partial sum of the infinite series $ \sum_{n=1}^\infty \left(\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}\right) $, you evaluate the series by adding up the first four terms of the series.

1. Write out the terms of the series for $ n = 1, 2, 3, $ and $ 4 $:
   - Term for $ n = 1 $: $ \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{1+1}} = 1 - \frac{1}{\sqrt{2}} $
   - Term for $ n = 2 $: $ \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2+1}} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} $
   - Term for $ n = 3 $: $ \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{3+1}} = \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} $
   - Term for $ n = 4 $: $ \frac{1}{\sqrt{4}} - \frac{1}{\sqrt{4+1}} = \frac{1}{\sqrt{4}} - \frac{1}{\sqrt{5}} $

2. Add up the first four terms:
   $$ \left(1 - \frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}\right) + \left(\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}}\right) + \left(\frac{1}{\sqrt{4}} - \frac{1}{\sqrt{5}}\right) $$

3. Simplify the terms and combine like terms:
   $$ 1 - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} - \frac{1}{2} + \frac{1}{2} - \frac{1}{\sqrt{5}} $$

4. Notice that many terms cancel out, leaving only $ 1 - \frac{1}{\sqrt{5}} $ as the fourth partial sum.

So, the fourth partial sum of the given infinite series is $ 1 - \frac{1}{\sqrt{5}} $.