Question

More sigma series questions (but for sums this time)!

Answer 1

I need to find the sum of \[\sum_{k=2}^{\infty} \frac{ 3 }{ k^2-k }\] and \[\sum_{k=0}^{\infty} \frac{ 1-3^k }{ 7^k }\] and I have no idea where to begin with simplifying these.

Answer 2

you can split into partial fractions in the first case.

or compare to 3/k^2-(1/100)k^2

(after some jth term it will be always greater)

Answer 3

The concept here is that once series converges after some jth term (say 1000000th), then it would converge starting from 1st term as well right?

Answer 4

( because the infinite series from jth term, is just lacking a finite scalar - i.e series from 1st to jth term)

Answer 5

am I being ridiculous?

Answer 6

HI!!

Answer 7

Not gonna lie, I did not understand half of that, m8.

Answer 8

second one or first?

Answer 9

Any help with either is great! I feel like these problems are JUST within reach but the stupid things are evading me like woah.

Answer 10

second one is easiest for sure \[\sum \frac{1}{7^k}-\sum \left(\frac{3}{7}\right)^k\] both are easy to add

Answer 11

you can do it in your head

Answer 12

\[\large \sum_{n=1}^{\infty}{\rm A}_n=\sum_{n=j}^{\infty}{\rm A}_n+\sum_{n=1}^{j-1}{\rm A}_n\]

is this correct?

Answer 13

what is \(1-\frac{1}{7}\)?

Answer 14

?

Answer 15

@misty1212 See, it seemed that way to me too, but the answer choices I have listed for this problem are: \[-\frac{ 7 }{ 6 }, -\frac{ 7 }{ 8 }, -\frac{ 7 }{ 18 }, -\frac{ 7 }{ 4 }, and - \frac{ 7 }{ 12 }.\]

Answer 16

we are not done yet

Answer 17

we haven't got the answer, we are working towards it
what is \(1-\frac{1}{7}\)?

Answer 18

@misty1212 Ah! Well then, it's just \[\frac{ 6 }{ 7 }\].

Answer 19

\[\large \sum_{n=1}^{\infty}{\rm A}_n=\sum_{n=j}^{\infty}{\rm A}_n+\sum_{n=1}^{j-1}{\rm A}_n\]

we know that
\[\sum_{n=1}^{j-1}{\rm A}_n\]is constant/scalar, so we can say

let\[k=\sum_{n=1}^{j-1}{\rm A}_n\]
so k is constant.

thus,
\[\large \sum_{n=1}^{\infty}{\rm A}_n=\sum_{n=j}^{\infty}{\rm A}_n+k\]

and this way if infinite series that starts from n=j converges, then the infinite series that starts from n=1 must converge (provided all terms from n=1 to n=j are defined).

(THEORY THAT IF SERIES CONVERGES AFTER JTH TERM THEN IT CONVERGES AFTER 1ST TERM)

Answer 20

it is going to take a couple steps
we have to compute each
then subtract
but both you can compute in your head
where \(\frac{6}{7}\) right
ans what is the reciprocal of \(\frac{6}{7}\)?

Answer 21

I know your questions are to evaluate actually, as opposed to determine convergence, but what I say is just for future to nail the series with comparison test.

Answer 22

i meant "and what is the reciprocal of \(\frac{6}{7}\)?

Answer 23

@fbi2015 Any bit helps, believe me. I do recognize that equation.

Answer 24

Reciprocal is \[\frac{ 7 }{ 6 }\]

Answer 25

we are gong to compute \[\sum\left(\frac{1}{7}\right)^k\] mentally \[1-\frac{1}{7}=\frac{6}{7}\] and the reciprocal of \(\frac{6}{7}=\frac{7}{6}\) so \[\sum\left(\frac{1}{7}\right)^k=\frac{7}{6}\]

Answer 26

repeat for \[\sum \left(\frac{3}{7}\right)^k\] \[1-\frac{3}{7}=\frac{4}{7}\] reciprocal is \(\frac{7}{4}\) so the sum is \(\frac{7}{4}\)

Answer 27

final answer \[\frac{7}{6}-\frac{7}{4}\] which you may be able to do in your head of not, but that is the last step

Answer 28

\[\frac{ 7 }{ 6 }-\frac{ 7 }{ 4 }=\frac{ 28 }{ 24 }-\frac{ 42 }{ 24 }=-\frac{ 7 }{ 12 }\]

Answer 29

And the reciprocals for each part of this equation are taken because everything is to the power of k?

Answer 30

lol no

Answer 31

it is because \[\sum r^n=\frac{1}{1-r}\] if \(|r|<1\)

Answer 32

FFFF that's right. ...Wow, I have that equation right next to me. WAY TO BE.

Answer 33

just saying you can compute it in your head because it is easy to subtract a number from 1, and easy to take the reciprocal

Answer 34

which is what \[\frac{1}{1-r}\] means

Answer 35

the first one requires more work unfortunately

Answer 36

any ideas or no? (just asking)

Answer 37

1st one partial fractions

Answer 38

telescoping series

Answer 39

the partial fractions for #1 are:

\[\sum_{k=2}^{\infty} \left(\frac{3}{k-1}-\frac{3}{k}\right)\]

Answer 40

\[\sum_{k=2}^{\infty} \left(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+\frac{3}{3}-\frac{3}{4}\right)=3-\lim_{k\to \infty}\frac{3}{k}=3-0=3\]

Answer 41

this is my first take

Answer 42

or, same way

\[\sum_{k=2}^{\infty} \left(\frac{3}{k-1}-\frac{3}{k}\right)=\sum_{k=2}^{\infty} \left(\frac{3}{k-1}\right)-\sum_{k=2}^{\infty}\left(\frac{3}{k}\right)=\sum_{k=1}^{\infty} \left(\frac{3}{k}\right)-\sum_{k=2}^{\infty}\left(\frac{3}{k}\right)=A_1=3\]

Answer 43

another way is to see that all the terms get killed off except for the three
plus, minus, plus, minus etc

Answer 44

I got \[\sum_{k=2}^{\infty}\frac{ 3 }{ k }-\frac{ 3 }{ k-1 }\]

Answer 45

yup

Answer 46

I suppose it doesn't matter because it'll telescope either way?

Answer 47

I don't know what you precisely mean by this

Answer 48

I proposed to approaches:

\[\sum_{k=2}^{\infty} \left(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+\frac{3}{3}-\frac{3}{4}\right)=3-\lim_{k\to \infty}\frac{3}{k}=3-0=3\]

and #2,

or, same way

\[\sum_{k=2}^{\infty} \left(\frac{3}{k-1}-\frac{3}{k}\right)=\sum_{k=2}^{\infty} \left(\frac{3}{k-1}\right)-\sum_{k=2}^{\infty}\left(\frac{3}{k}\right)=\sum_{k=1}^{\infty} \left(\frac{3}{k}\right)-\sum_{k=2}^{\infty}\left(\frac{3}{k}\right)=A_1=3\]

Answer 49

my notations were off a bit, but you get the concept

Answer 50

Your equation was a flipped version of mine is what I meant. And yeah, I get it. Thanks!

Answer 51

Good! Cookies!