Question

find the sum of the series

Answer 1

\[\sum_{n=1}^{\infty} \frac{ 3 }{ n(n+1) }+\frac{ 1 }{ 2n }\]

Answer 2

i separated it them \[\sum_{n=1}^{\infty} \frac{ 2 }{ n(n+1)} + \sum_{n=1}^{\infty} \frac{ 1 }{ 2^n }\]

Answer 3

and i dont know what to do from here

Answer 4

some difference between 1st and 2nd in \(\dfrac{1}{2n}\) and \(\dfrac{1}{2^n}\)

Answer 5

i forgot to put the n in exponent form

Answer 6

i was trying to type it as fast i possibly could

Answer 7

so which one is correct

Answer 8

the \[\frac{ 1 }{ 2^n }\]

Answer 9

second sum is geometric series

Answer 10

how you know that?

Answer 11

if you expand you get

1/2, 1/2^2, 1/2^3, ...

geometric with common ratio 1/2

Answer 12

ok so first try to work with


\(\large \color{black}{\begin{align}
\sum_{n=1}^{\infty} \frac{ 2 }{ n(n+1)} \hspace{.33em}\\~\\
\end{align}}\)

for that u need to split the term in \(\Large \text{partial fractions}\)

Answer 13

oh okay i got it to be \[\frac{ 3 }{ n }-\frac{ 3 }{ n+1}\]

Answer 14

is it right?

Answer 15

i think it had \(2\) instead of \(3\)

Answer 16

lol its \[\frac{ 3 }{ n(n+1)}\]

Answer 17

sorry about that

Answer 18

\(\large \color{black}{\begin{align}
\sum_{n=1}^{\infty} \frac{ 3 }{ n(n+1)} \hspace{.33em}\\~\\
\implies 3\sum_{n=1}^{\infty} \frac{ 1 }{ n(n+1)} \hspace{.33em}\\~\\
\implies 3\sum_{n=1}^{\infty} \frac{ 1+n-n }{ n(n+1)} \hspace{.33em}\\~\\
\implies 3\sum_{n=1}^{\infty} \frac{ 1+n}{ n(n+1)}-\frac{n}{ n(n+1)} \hspace{.33em}\\~\\
\implies 3\sum_{n=1}^{\infty} \frac{ 1}{ n}-\frac{1}{ (n+1)} \hspace{.33em}\\~\\
\end{align}}\)

Answer 19

ok so u have now \(3\) outside .

now write the with each term of series upto \( n=6\)

\(\large \color{black}{\begin{align}

\implies \sum_{n=1}^{6} \frac{ 1}{ n}-\frac{1}{ (n+1)} \hspace{.33em}\\~\\
\end{align}}\)

Answer 20

okay

Answer 21

we would be left with \[\frac{ 1 }{ 1 } - \frac{ 1 }{ 7 }\]

Answer 22

\(\large \color{black}{\begin{align}

\implies& \sum_{n=1}^{4 } \frac{ 1}{ n}-\frac{1}{ (n+1)} \hspace{.33em}\\~\\
\implies& \frac{ 1}{ 1}-\frac{1}{ (1+1)}+ \frac{ 1}{ 2}-\frac{1}{ (2+1)} +\frac{ 1}{ 3}-\frac{1}{ (3+1)}+ \frac{ 1}{ 4}-\frac{1}{ (4+1)}\hspace{.33em}\\~\\
\implies &\frac{ 1}{ 1}-\frac{1}{ 2}+ \frac{ 1}{ 2}-\frac{1}{3} +\frac{ 1}{ 3}-\frac{1}{4}+ \frac{ 1}{ 4}-\frac{1}{ 5}\hspace{.33em}\\~\\
\implies & 1-\dfrac15
\end{align}}\)

if this goes till \(\infty\) then

\(\large \color{black}{\begin{align}

\implies& \sum_{n=1}^{\infty } \frac{ 1}{ n}-\frac{1}{ (n+1)} \hspace{.33em}\\~\\

\implies& 1-\dfrac{1}{\infty} \hspace{.33em}\\~\\
\implies& 1 \hspace{.33em}\\~\\
\end{align}}\)

Answer 23

mathmath333 is using a telescoping series

http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/SeriesTests/telescoping.html

Answer 24

so for the n exponent it would just be (1/2)^1 + (1/2)^2 + (1/2)^3+....(1/2)^infinity?

Answer 25

yes it is one of the telescooping series in which the series terms cancel itself.

so

so

\(\large \color{black}{\begin{align}

\implies& 3\sum_{n=1}^{\infty } \left
(\frac{ 1}{ n}-\frac{1}{ (n+1)}\right)=3 \hspace{.33em}\\~\\

\end{align}}\)

Answer 26

for the \[\frac{ 1 }{ 2^n }\]

Answer 27

you would do the same thing?

Answer 28

\(\large \color{black}{\begin{align}

y&=\sum_{n=1}^{\infty } \frac{ 1}{ 2^n} \hspace{.33em}\\~\\

y&=\frac{ 1}{ 2^1}+ \frac{ 1}{ 2^2}+\frac{ 1}{ 2^3}+\frac{ 1}{ 2^4}\cdots \infty \hspace{.33em}\\~\\
y&=\frac{ 1}{ 2^1}\left(1+ \frac{ 1}{ 2^1}+\frac{ 1}{ 2^2}+\frac{ 1}{ 2^3}\cdots \infty\right) \hspace{.33em}\\~\\
y&=\frac{ 1}{ 2}\left(1+ y\right) \hspace{.33em}\\~\\
2y&=1+ y \hspace{.33em}\\~\\
2y-y&=1 \hspace{.33em}\\~\\
y&=1 \hspace{.33em}\\~\\
& \Large \sum_{n=1}^{\infty } \frac{ 1}{ 2^n}=1

\end{align}}\)

Answer 29

see if it makes sense

Answer 30

could you distribute the 1/2?

Answer 31

or not?

Answer 32

\(\large \color{red}{\begin{align}

y&=\frac{ 1}{ 2}\left(1+ y\right) \hspace{.33em}\\~\\


\end{align}}\)

u mean here

Answer 33

yes

Answer 34

multiply both sides by \(\Large 2\)

Answer 35

you do that every time theres a 1/2?

Answer 36

\(\large \color{red}{\begin{align}

y&=\frac{ 1}{ 2}\left(1+ y\right) \hspace{.33em}\\~\\
y&=\frac{ 1}{ 2}+ \frac{ y}{ 2} \hspace{.33em}\\~\\

\end{align}}\)

we can e distribute the \(\dfrac12\) either way u will get the same answer

Answer 37

okay its just that every time i see something like i think of distributing it for some reason

Answer 38

ok so u know the answer now

Answer 39

yes thanks again @mathmath333