How do you find a formula for the nth partial sum of the series [5/1*2]+[5/2*3]+[5/3*4]+...+[5/n(n+1)]+... and use it to find the series' sum if the series converges?

Answer 1

To find a formula for the nth partial sum of the series ( \frac{5}{n(n+1)} ), we can first write out a few terms of the series and observe a pattern. Then we can use this pattern to derive a general formula for the nth partial sum.

The series is:

[ S_n = \frac{5}{1 \cdot 2} + \frac{5}{2 \cdot 3} + \frac{5}{3 \cdot 4} + \ldots + \frac{5}{n(n+1)} ]

To find the sum of this series, if it converges, we can use the formula for the nth partial sum and take the limit as ( n ) approaches infinity. If this limit exists, it represents the sum of the series.

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Answer 2
First of all we can factor the #5#:
#sum_1^(+oo)5/(n(n+1))=5sum_1^(+oo)1/(n(n+1))=(1)#
This series, except for the factos #5#, is named Mengoli Series.
Now let's try to separate the term #1/(n(n+1)# in the sum of two terms:
#1/(n(n+1))=A/n+B/(n+1)=(A(n+1)+Bn)/(n(n+1))=#
#=(n(A+B)+B)/(n(n+1)#.

For the identity principle of polynomials:

#1=n(A+B)+B#

So:

#A+B=0# #A=1#

That gives:

#A=1# #B=-1#.

So:

#(1)=5sum_1^(+oo)(1/n-1/(n+1))#.
Let's see now #S_n#:
#S_n=5(1/1-1/2+1/2-1/3+1/3-1/4+...+1/n-1/(n+1))#,

We can notice now that all the terms except the first and the last will erase themselves, so:

#S_n=5(1-1/(n+1))# and we know that we can obtain the sum of ALL the terms of the series with this limit:
#lim_(nrarr+oo)S_n=lim_(nrarr+oo)5(1-1/(n+1))=5(1-0)=5#.
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Answer from HIX Tutor

When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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