How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #1/(1*3)+1/(3*5)+1/(5*7)+...+1/((2n-1)(2n+1))+...#?

Question

Paisley Johnson · Accepted Answer

undefined To find the nth partial sum of the series $ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2n-1)(2n+1)} + \ldots $, we first express each term as a fraction and then find a pattern in the series.

The nth term of the series can be written as:
$$ \frac{1}{(2n-1)(2n+1)} = \frac{1}{4n^2 - 1} $$

Now, let's find the nth partial sum, denoted as $ S_n $:
$$ S_n = \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2n-1)(2n+1)} $$

$$ S_n = \frac{1}{4n^2 - 1} + \frac{1}{4(n-1)^2 - 1} + \frac{1}{4(n-2)^2 - 1} + \ldots + \frac{1}{1^2 - 1} $$

Now, let's simplify the terms in $ S_n $ and see if there's a pattern:
$$ S_n = \frac{1}{4n^2 - 1} + \frac{1}{4n^2 - 9} + \frac{1}{4n^2 - 25} + \ldots + \frac{1}{0} $$

$$ S_n = \frac{1}{4n^2 - (2n+1)^2} + \frac{1}{4n^2 - (2n+3)^2} + \frac{1}{4n^2 - (2n+5)^2} + \ldots + \frac{1}{0} $$

We can observe that each term in the series can be written in the form $ \frac{1}{a^2 - b^2} $, which can be decomposed using partial fraction decomposition. However, there isn't a simple arithmetic pattern in the series to directly find a closed-form expression for $ S_n $.

To determine whether the series converges, we can examine the behavior of the nth term as $ n $ approaches infinity. Notice that as $ n $ becomes very large, the terms in the series approach zero. Therefore, by the nth-term test for divergence, since the limit of the nth term as $ n $ approaches infinity is zero, the series converges.

However, finding the exact sum when it exists requires more advanced techniques such as partial fraction decomposition or using other methods like telescoping series.

Luke Alvarez · Answer

The nth partial sunm is #=n/(2n+1)#
The series converge to #1/2# Let's perform the partial fraction decomposition of the nth term #1/((2n-1)(2n+1))=A/(2n-1)+B/(2n+1)# #=(A(2n+1)+B(2n-1))/((2n-1)(2n+1))# The numerators are the same, we compare the denominators #1=A(2n+1)+B(2n-1))# Let #n=1/2#, #=>#, #1=2A#, #=>#, #A=1/2# Let #n=-1/2#, #=>#, #1=-2B#, #=>#, #B=-1/2# Therefore, #1/((2n-1)(2n+1))=(1/2)/(2n-1)-(1/2)/(2n+1)# #u_n=1/2(1/(2n-1)-1/(2n+1))# So, #u_1=1/2(1/1-cancel(1/3))# #u_2=1/2(cancel(1/3)-cancel(1/5))# #u_3=1/2(cancel(1/5)-cancel(1/7))# #u_(n-1)=1/2(cancel(1/(2n-3))-cancel(1/(2n-1)))# #u_n=1/2(cancel(1/(2n-1))-1/(2n+1))# #sum_(n=1) ^n=1/2(1-1/(2n+1))=1/2((2n)/(2n+1))# #=n/(2n+1)# The nth partial sunm is #=n/(2n+1)# #lim_(n->+oo)n/(2n+1)=lim_(n->+oo)n/(2n)=1/2# The series converge to #1/2#