Question

for the following telescoping series find a formula for the nth term of the sequence of partial sums {Sn or S subscript n}. Then evaluate lim of S subscript n (Sn) as n->infinity to obtain value of the series or that the series diverges

sum_{k=1}^{infty} 1/(ak+1)(ak+a+1)

Answer 1

\[\sum_{k=1}^{\infty} 1/(ak+1)(ak+a+1) \]

Answer 2

my guess is you will see that it telescopes by partial fraction decomposition
that is what you need to try

Answer 3

yeah i need help coming up with the partial fraction

Answer 4

first notice that this can be also written as
\[\sum_{k=1}^{\infty}(1/a(ak+1))-(1/a(ak+a+1))\]
\[=\sum_{k=1}^{\infty}1/a(1/(ak+1))-(1/(ak+a+1))\]
\[=1/a(\sum_{k=1}^{\infty}(1/(ak+1))-(1/(ak+a+1)))\]
so now let's look at Sn which is
\[s_n=1/a(\sum_{k=1}^{n}(1/(ak+1))-(1/(ak+a+1)))\]
\[s_n=1/a(\sum_{k=1}^{\infty}((ak+1))-(1/(a(k+1)+1)))\]
now if we are to expand this we'll have:
\[1/a(1/(a+1)-1/(2a+1)+1/(2a+1)+...-1/(an+1)+1/(an+1)\]
\[-1/(a(n+1)+1))\]
as you can see, the terms between 1/(a+1) and -1/(a(n+1)+1) will be cancelled out except for 1/(a+1) and -1/(a(n+1)+1) themselves. if that's the case, this expanded form would then end up to
\[1/a(1/(a+1)-1/(a(n+1)+1))=1/(a^2+a)-1/(a^2n+a^2+a)\]
so our Sn is nothing but:
\[s_n=1/(a^2+a)-1/(a^2n+a^2+a)\]
now let us evaluate lim of S subscript n (Sn) as n->infinity to obtain value of the series:
\[\lim_{n \rightarrow \infty} s_n=\lim_{n \rightarrow \infty}1/(a^2+a)-1/(a^2n+a^2+a)=1/(a^2+a)-0\]
\[=1/(a^2+a)\]

Answer 5

please disregard the equation:
\[s_n=1/a(\sum_{k=1}^{\infty}((ak+1))-(1/a(k+1)+1)))\]
my fingers slipped

Answer 6

okay i'm writing this on paper to understand what is going on. i'm kind of lost in the beginning sorry. give me a sec

Answer 7

ah your confused at the partial fraction decomposition part?

Answer 8

yes

Answer 9

ahhh this is so confusing. i'll just bring this to my prof. thanks for your help though. your answer look right. i'm just bad with math

Answer 10

@mrlunarugby some parts of the equation I wrote have been cutted since some are too long to be displayed

Answer 11

@mrlunarugby being good at math just needs dedication :D Its not all about intelligence