Infinite Series

[intervade]

Ok, so I'm sort of understanding this problem but not all the way.

If the nth partial sum of a series
\(\displaystyle \sum_{n=1}^{\infty }a_{n}\)

is
\(\displaystyle s_{n} = \frac{n}{3(n+3)}\)

find
\(\displaystyle a_{n}\) and \(\displaystyle \sum_{n=1}^{\infty }a_{n}\)

Now, I started going with s ..
\(\displaystyle s_{1} = \frac{1}{12}\), \(\displaystyle s_{2} = \frac{2}{15}\), \(\displaystyle s_{3} = \frac{3}{18}\), \(\displaystyle s_{4} = \frac{4}{21}\) and \(\displaystyle s_{5} = \frac{5}{24}\)

From this I've concluded that
\(\displaystyle a_{1} = \frac{1}{12}\), \(\displaystyle a_{2} = \frac{1}{20}\), \(\displaystyle a_{3} = \frac{1}{30}\), \(\displaystyle a_{4} = \frac{1}{42}\) and \(\displaystyle a_{5} = \frac{1}{56}\)

first of all, does this get me anywhere? Am I on the right track? How do I find my nth term \(\displaystyle a_{n}\)?

Help on this would be much appreciated!

 

[Deleted member 4993]

Ok, so I'm sort of understanding this problem but not all the way.

If the nth partial sum of a series
\(\displaystyle \sum_{n=1}^{\infty }a_{n}\)

is
\(\displaystyle s_{n} = \frac{n}{3(n+3)}\)

find
\(\displaystyle a_{n}\) and \(\displaystyle \sum_{n=1}^{\infty }a_{n}\)

Now, I started going with s ..
\(\displaystyle s_{1} = \frac{1}{12}\), \(\displaystyle s_{2} = \frac{2}{15}\), \(\displaystyle s_{3} = \frac{3}{18}\), \(\displaystyle s_{4} = \frac{4}{21}\) and \(\displaystyle s_{5} = \frac{5}{24}\)

From this I've concluded that
\(\displaystyle a_{1} = \frac{1}{12}\), \(\displaystyle a_{2} = \frac{1}{20}\), \(\displaystyle a_{3} = \frac{1}{30}\), \(\displaystyle a_{4} = \frac{1}{42}\) and \(\displaystyle a_{5} = \frac{1}{56}\)

first of all, does this get me anywhere? Am I on the right track? How do I find my nth term \(\displaystyle a_{n}\)?

Help on this would be much appreciated!

for

\(\displaystyle \sum_{n=1}^{\infty }a_{n} = \lim_{n to \infty} s_n\)

and

\(\displaystyle a_n = s_n - s_{n-1}\)

 

[soroban]

Hello, intervade!

Ok, so I'm sort of understanding this problem but not all the way.

\(\displaystyle \text{If the }n^{th}\text{ partial sum of a series }\sum_{n=1}^{\infty }a_{n}\,\text{ is: }\;S_n \:=\: \frac{n}{3(n+3)}\)
\(\displaystyle \text{Find: }\;a_n \:\text{ and }\:\sum_{n=1}^{\infty }a_{n}\)

\(\displaystyle \text{I started with: }\;\;\begin{Bmatrix}S_1 &=& \frac{1}{12} \\ \\[-3mm] S_2 &=& \frac{2}{15}\\ \\[-3mm] S_3 &=& \frac{3}{18} \\ \\[-3mm] S_4 &=& \frac{4}{21} \\ \\[-3mm] S_5 &=& \frac{5}{24} \\ \vdots && \vdots \end{Bmatrix}\)


\(\displaystyle \text{From this I've concluded that: }\;\begin{Bmatrix}a_1 &=& \frac{1}{12} \\ \\[-3mm] a_2 &=& \frac{1}{20} \\ \\[-3mm] a_3 &=& \frac{1}{30} \\ \\[-3mm] a_4 &=& \frac{1}{42} \\ \\[-3mm] a_5 &=& \frac{1}{56} \\ \vdots && \vdots \end{Bmatrix}\)


First of all, does this get me anywhere? .Am I on the right track? . . Yes, definitely!

\(\displaystyle \text{Note that: }\;\begin{Bmatrix} \frac{1}{12} &=& \frac{1}{3\cdot4} \\ \\[-3mm] \frac{1}{20} &=& \frac{1}{4\cdot5} \\ \\[-3mm] \frac{1}{30} &=& \frac{1}{5\cdot6} \\ \\[-3mm] \frac{1}{42} &=& \frac{1}{6\cdot7} \\ \vdots && \vdots \end{Bmatrix}\)

\(\displaystyle \text{Therefore, the }n^{th}\text{ term is: }\;\boxed{a_n\;=\;\frac{1}{(n+2)(n+3)}}\)



\(\displaystyle \text{We are told that: }\;S_n \;=\;\frac{n}{3(n+3)}\)

\(\displaystyle \text{Then: }\;\sum^{\infty}_{n=1}a_n \;=\;\lim_{n\to\infty}\frac{n}{3(n+3)} \;=\;\lim_{n\to\infty}\frac{n}{3n+9}\)


\(\displaystyle \text{Divide top and bottom by }n\!:\;\;\lim_{n\to\infty}\,\frac{\frac{n}{n}}{\frac{3n}{n} + \frac{9}{n}} \;=\; \lim_{n\to\infty}\frac{1}{3 + \frac{9}{n}} \;=\;\frac{1}{3+0} \;=\;\frac{1}{3}\)

\(\displaystyle \text{Therefore: }\;\sum^{\infty}_{n=1} a_n \;=\;\frac{1}{3}\)