Question

limit as n approach infinity for the summation of 1/[(n+3)(n+4)] from n=1 to infinity

Answer 1

I think you have written this funny

Answer 2

\[\lim_{n \rightarrow \infty} \sum_{n=1}^{ \infty}1/[(n+3)(n+4)]\]

Answer 3

is it
\[\sum_{n=1}^{\infty}\frac{1}{(n+3)(n+4)}\]?

Answer 4

ok gimmick is partial fractions
rewrite at
\[\frac{1}{n+3}-\frac{1}{n+4}\] and you will see that your sum "telescopes"

Answer 5

i want to know whats next after we partialise

Answer 6

observe how the terms behave, try writing out the first 3 or so...

Answer 7

ok sure, but first of all the thing you wrote above doesn't really make any sense
it is
\[\sum_{n=1}^{\infty}a_n:=\lim_{k\to \infty}\sum_{n=1}^ka_n\]

Answer 8

do what mr @TuringTest said. replace n by 1, by 2, by 3 etc and you will see exactly what happens
since the second term is minus the previous one, you will have a raft of cancellation, leaving only a term or two
check it

Answer 9

do you mean to say that the terms b4 partialising are different from those aftr partialising

Answer 10

what are the first three terms in the series after applying partial fractions?

Answer 11

1/20+1/30+1/42.......

Answer 12

no, do it in the way that satellite showed you in terms of the representation in partial fractions

Answer 13

\[\frac1{n+3}-\frac1{n+4}\]what are the first three terms?

Answer 14

...or however many you feel like writing

Answer 15

[1/4-1/5]+[1/5-1/6]+[1/6-1/7]+[1/7-1/8].......

Answer 16

nice, now are you seeing some things cancelling?

Answer 17

yes

Answer 18

what is left?

Answer 19

the first and the last term

Answer 20

but what is the last term as n goes to infinity?

Answer 21

it is infinity

Answer 22

nice try, but infinity is not a number, so the answer is that there \(is~no~last~term\)
for whatever term we call the last we can always make another that will cancel the one before it, hence they will \(all\) cancel except the first term

the logic here may be a bit counter-intuitive at first, but remember
1) infinity is not a number; it is a concept
2) there is no last term, we can always add another to cancel the one before it

Answer 23

...meditate on this, it will help you develop a mathematical intuition ;)

Answer 24

okay, so we are only left with the first term ?

Answer 25

yes

Answer 26

\[(\frac14-\cancel{\frac15})+(\cancel{\frac15}-\cancel{\frac16})+(\cancel{\frac16}-\cancel{\frac17})+...\]you can see the pattern of cancellation continues literally forever, so no last term remains, only the first

Answer 27

okay i get that

Answer 28

okay, so what does that tell you about\[\sum_{n=1}^\infty\frac1{(n+3)(n+4)}=\sum_{n=1}^\infty\frac1{n+3}-\frac1{n+4}\]?

Answer 29

that the limit approach approach the first term

Answer 30

no limit, the sum of the series is 1/4
(it's not the first term, which is 1/4-1/5)

Answer 31

now here is where your question makes no sense\[\sum_{n=1}^\infty\frac1{n+3}-\frac1{n+4}=\frac14\]but you typed the problem as\[\lim_{n\to\infty}\sum_{n=1}^\infty\frac1{n+3}-\frac1{n+4}=\lim_{n\to\infty}\frac14=\frac14\]so the limit thing is redundant and adds nothing to the problem
the sum of the series is a constant, and the limit of a constant is a constant, so why the hell is that even in there

if you made a typo that would make sense, for instance if it were....

Answer 32

\[\lim_{n\to\infty}\sum_{i=1}^n\frac1{i+3}-\frac1{i+4}=\lim_{n\to\infty}\frac14=\sum_{i=1}^\infty\frac1{i+3}-\frac1{i+4}=\frac14\]that would not be redundant as it would be talking about the limit of the highest term in the series n approaching infinity
same answer either way though, so who cares :P

Answer 33

typo that was meant to read\[\lim_{n\to\infty}\sum_{i=1}^n\frac1{i+3}-\frac1{i+4}=\sum_{i=1}^\infty\frac1{i+3}-\frac1{i+4}=\frac14\]

Answer 34

thnx