How do you find the 4th partial sum of the infinite series #sum_(n=1)^oo(1/sqrt(n)1/sqrt(n+1))# ?
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.
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by regrouping,
by cancelling intermediate terms,
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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.

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}} )

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) ]

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}} ]

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}} ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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