Question

If both the sequence (nan) and the series
n(an - an+1) converge, prove

that the series (an) also converges.

hint: consider the sequence of partial sums and use the ALT for sequences.

Answer 1

dizzy :D

Answer 2

The sequence \(\large na_n\) converges and the series \[\Large \sum_{n=k}^\infty n(a_n-a_{n+1})\] also converges... so what are we trying to prove again? ^.^

Answer 3

that (an) converges as well

Answer 4

as a sequence or as a series?

Answer 5

oo series sorry i left that part out of the ques

Answer 6

You want to prove that \[\Large \sum_{n=k}^\infty a_n\] is also convergent?

Answer 7

says with n=1

Answer 8

yes*

Answer 9

looks daunting.
but anyway, let's consider the sequence of partial sums ^.^
\[\Large s_n:=\Large \sum_{k=1}^nk(a_n-a_{n+1})\]

Answer 10

ok

Answer 11

whoops, i meant
\[\Large s_n:=\Large \sum_{k=1}^nk(a_k-a_{k+1})\]

Answer 12

np i understood what u meant

Answer 13

So, can you derive a nicer expression for \(\large s_n\)? One that doesn't involve that big sigma...

Answer 14

i dont follow?

Answer 15

Here, I'll show you... what's \(\large s_1\) equal to?

Answer 16

1?

Answer 17

or 0

Answer 18

no no...
Just substitute, n = 1
\[\Large s_1 = \sum_{k=1}^{1}k(a_k-a_{k+1})\]

Answer 19

or wait

Answer 20

I believe this is equal to \[\Large a_1 - a_2\]?

Answer 21

is it a1 - a2?

Answer 22

ok yes sorry

Answer 23

Okay, kick it up a notch ^.^
What about \(\large s_2\)?

Answer 24

2(a2-a3)

Answer 25

?

Answer 26

No, that's just the second term of the sequence, but if you consider it a series, you have to add it to the previous term...
\[\Large a_1 - a_2 + 2(a_2-a_3) \]

Answer 27

oo ok so then s3= a1-a2+2(a2-a3)+3(a3-a4)?

Answer 28

I think we can simplify a2 before we proceed to a3\[\Large a_1 - a_2 + 2(a_2-a_3) = a_1+a_2 - 2a_3\]

Answer 29

ok makes sense

Answer 30

Now you can add the third term. s3 = ?

Answer 31

a1+a2+a3-3a4?

Answer 32

That's right LOL

Answer 33

Can you now find a nice representation for sn?

Answer 34

sn=a1+a2+a3....+an

Answer 35

or ..+an+1?

Answer 36

<whistles>

Answer 37

Allow me to make a list of what we know ^_^
\[\Large s_1 = a_1 - a_2\]\[\Large s_2 = a_1+a_2 - 2a_3\]\[\Large s_3 = a_1+a_2+a_3-3a_4\]\[\Large s_4 = a_1+a_2+a_3+a_4-4a_5\]
and so on, right?

Answer 38

oo right

Answer 39

im drawing a blank on how to put it into one summation

Answer 40

Well, how about this...\[\Large s_1 = a_1+a_2 -2a_2\]\[\Large s_2= a_1+a_2+a_3 - 3a_3\]\[\Large s_3 = a_1+a_2+a_3 + a_4 -4a_4\]\[\Large s_4 = a_1+a_2+a_3+a_4 +a_4 - 5a_5\]
A little modification.

Answer 41

sorry, typo on the last part
\[\Large s_4 = a_1+a_2+a_3+a_4 +a_\color{red}5 - 5a_5\]

Answer 42

ok i was just gunna ask that lol

Answer 43

As for compressing them into one summation, you're severely lacking imagination XD

Answer 44

\[\Large s_{n-1} = \left[\sum_{k=1}^{n}a_k\right]-na_n\]

Answer 45

ooo ok got it

Answer 46

But you know sn is convergent right?

Answer 47

correct

Answer 48

Because it's the series of partial sums

Answer 49

and so is nan

Answer 50

We can fix this so that it becomes
\[\Large s_{n-1}+na_n=\sum_{k=1}^na_k \]

Answer 51

That was just algebra... are you still with me? :)

Answer 52

yes sorry im just looking over the steps as we are going

Answer 53

oh okay take your time ^.^

Answer 54

ok go it

Answer 55

got*

Answer 56

This part says it all:
\[\Large s_{n-1}+na_n=\sum_{k=1}^na_k\]

You know THIS converges (this is simply the sequence of partial sums of a known convergent series)
\[\Large \color{green}{s_{n-1}}+na_n=\sum_{k=1}^na_k\], so suppose it converges to some value s



You also know that this converges (it was given in your question) suppose it converges to some value x.\[\Large s_{n-1}+\color{red}{na_n}=\sum_{k=1}^na_k\]

Answer 57

So if we take the limit on both sides as n goes to infinity...
\[\Large \lim_{n\rightarrow \infty}\left (s_{n-1}+na_n\right)=\lim_{n\rightarrow\infty }\sum_{k=1}^na_k\]
We get
\[\Large \color{green}s+\color{red}x=\color{blue}{\sum_{k=1}^\infty a_k}\]

Answer 58

I wouldn't mind a little applause ^.^

Answer 59

lol ur a life saver!..i never understand this stuff but when u take me thru it step by step i get it more..thank you soo much!

Answer 60

:D

Answer 61

i have another question if u have time lol

Answer 62

This was a mouthful. go ahead :D

Answer 63

this one is even harder lol..ok il close this one and post another