Question

See comments, please halp. :C

Answer 1

Um, I believe the question messed up, retype it again.

Answer 2

Could anyone give a clue on how to rewrite \[\sum_{k=1}^{2(m+1)}\frac{(-1)^{k+1}}{k}=\sum_{k=m+2}^{2(m+1)}\frac{1}{k}\] given that you know \[\sum_{k=1}^{2m}\frac{(-1)^{k+1}}{k}=\sum_{k=m+1}^{2m}\frac{1}{k}\]

Answer 3

I'm thinking of changing the "interval" of the left term sum so that you can utilize and rewrite as the right term in the second statement.

Answer 4

You're still left with \[\sum_{k=2m}^{2(m+2)} \frac{(-1)^{k+1}}{k}\] though and I'm not sure where to go from here.

Answer 5

O_o

Answer 6

Is it possible to perhaps "bake" this together with \[\sum_{k=m+1}^{2m}\frac{1}{k}\], if so how? Not sure if this is a step in the right direction.

Answer 7

are you solving for \(m\)?

Answer 8

It's not a solve I'm looking for, I'm trying to prove by induction that the statement is true.

Answer 9

It shouldn't be that hard tbh. :(

Answer 10

oh what is the original statement?

Answer 11

Maybe my initial induction proposition is clumsily chosen(?) -- it's the second one in the first comment.

Answer 12

oh i see, you are going from \(m\) to \(m+1\) i get it, but somehow this looks kind of unlikely

Answer 13

Prove that \[\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}=\sum_{k=n+1}^{2n}\frac{1}{k}, \forall n \in\]

Answer 14

wow there is a fact i did not know

Answer 15

Belongs to natural that is.

Answer 16

It's quite a basic proof also which bothers the hell out of me, been sitting with this for some time. >_<

Answer 17

let me right it on paper and see if i can do it
usually summations are easy because the previous case stares you right in the face

Answer 18

That's what I think is happening, which is even more cruel, haha.

Answer 19

Maybe it's easier to try to rewrite the right term instead? (When you're trying to prove for n=m+1)

Answer 20

just a sec, it think it is just algebra

Answer 21

you got it?

Answer 22

nope

Answer 23

I was thinking that might be a way to go about solving it, but it didn't feel like it.

Answer 24

ok i think we can do it easily, just give me a second
in fact, i don't think it is algebra at all

Answer 25

we know this \[\sum_{k=1}^{2m}\frac{(-1)^{k+1}}{k}=\sum_{k=m+1}^{2m}\frac{1}{k}\] we want this \[\sum_{k=1}^{2(m+1)}\frac{(-1)^{k+1}}{k}=\sum_{k=m+2}^{2(m+1)}\frac{1}{k}\]

Answer 26

I'm quite certain it's about utilizing the "range" of the sum (not sure if you know what I mean by range, not native eng. )

Answer 27

the left hand side can be rewritten as
\[\sum_{k=1}^{2(m+1)}\frac{(-1)^{k+1}}{k}=\sum_{k=1}^{2m}\frac{(-1)^{k+1}}{k}+\frac{1}{2m+1}-\frac{1}{2m+2}\]

Answer 28

Oh, that's right. Still too rusty with sums. :/

Answer 29

if my algebra is correct, those are just the last two terms of the sum

Answer 30

so by induction
\[\sum_{k=1}^{2m}\frac{(-1)^{k+1}}{k}+\frac{1}{2m+1}-\frac{1}{2m+2}=\sum_{k=m+1}^{2m}\frac{1}{k}+\frac{1}{2m+1}-\frac{1}{2m+2}\]

Answer 31

Thing is, how do you connect/rewrite that into a single sum with initial element k=m+2 and last element 2m+2?

Answer 32

I don't see it. :(

Answer 33

give me a sec
all we have to do is write the difference between \[\sum_{k=m+2}^{2(m+1)}\frac{1}{k}\] and \[\sum_{k=m+1}^{2m}\frac{1}{k}\]

Answer 34

to get from the bottom one to the top one, you have two extra terms at the end, minus the first term
\[\frac{1}{2m+1}+\frac{1}{2m+2}-\frac{1}{m+1}\]

Answer 35

now if by some miracle
\[\frac{1}{2m+1}-\frac{1}{2m+2}=\frac{1}{2m+1}+\frac{1}{2m+2}-\frac{1}{m+1}\] then we would have it

Answer 36

Don't you have to "add" that last term seeing as how the above sum doesn't "include" it? Or am I thinking completely wrong

Answer 37

you want to get from\[\sum_{k=m+1}^{2m}\frac{1}{k}+\frac{1}{2m+1}-\frac{1}{2m+2}\] to \[\sum_{k=m+2}^{2(m+1)}\frac{1}{k}\]

Answer 38

and \[\sum_{k=m+2}^{2(m+1)}\frac{1}{k}=\sum_{k=m+1}^{2m}\frac{1}{k}+\frac{1}{2m+1}+\frac{1}{2m+2}-\frac{1}{m+1}\]

Answer 39

lets see if the algebra works

Answer 40

Hey, is there any way I can click, like... a "leave" button because I'm getting a lot of unwanted notifications..

Answer 41

yes, it works!~!!!

Answer 42

I don't see how you derive to that last statement.

Answer 43

this one?\[\sum_{k=m+2}^{2(m+1)}\frac{1}{k}=\sum_{k=m+1}^{2m}\frac{1}{k}+\frac{1}{2m+1}+\frac{1}{2m+2}-\frac{1}{m+1}\]

Answer 44

Yes! :C

Answer 45

the only difference between the first sum and the second sum is:
the first one goes up to \(2m+2\) instead of just \(2m\) so it inludes \(\frac{1}{2m+1}\) and \(\frac{1}{2m+2}\) whereas the second sum does not

Answer 46

also the first one starts at \(m+2\) whereas the second one starts at \(m+1\)
so the first one is missing the \(\frac{1}{m+1}\) term

Answer 47

that is why \[\sum_{k=m+2}^{2(m+1)}\frac{1}{k}=\sum_{k=m+1}^{2m}\frac{1}{k}+\frac{1}{2m+1}+\frac{1}{2m+2}-\frac{1}{m+1}\] it has nothing to do with the induction, it is just a fact

Answer 48

Ooh, I see! But how did you arrive to \[\sum_{k=m+2}^{2(m+1)}\frac{1}{k}\] from the initial left hand side? (also; I don't know whalexnuker)

Answer 49

ok lets start at the beginning so we can clean this up
okay?

Answer 50

Yes!

Answer 51

we know \[\sum_{k=1}^{2(m+1)}\frac{(-1)^{k+1}}{k}=\sum_{k=1}^{2m}\frac{(-1)^{k+1}}{k}+\frac{1}{2m+1}-\frac{1}{2m+2}\] just a fact, tack on the last two terms

Answer 52

it is taking me a while to write, give me a sec

Answer 53

Yes, don't worry; your help is extremely appreciated. Thanks for taking the time. :)

Answer 54

by induction, this you can replace
\[\sum_{k=1}^{2m}\frac{(-1)^{k+1}}{k}\] by \[\sum_{k=m+1}^{2m}\frac{1}{k}\]
giving \[\sum_{k=1}^{2(m+1)}\frac{(-1)^{k+1}}{k}=\sum_{k=m+1}^{2m}\frac{1}{k}+\frac{1}{2m+1}-\frac{1}{2m+2}\]

Answer 55

by algebra \[\frac{1}{2m+1}-\frac{1}{2m+2}=\frac{1}{2m+1}+\frac{1}{2m+2}-\frac{1}{m+1}\]

Answer 56

this you can check by doing the computation and seeing you get the same answer

Answer 57

._. Thanks for ignoring me.

Answer 58

There is no way that I'm aware of Whalexnuker.

Answer 59

and then by writing as a sum, we see \[\sum_{k=1}^{2(m+1)}\frac{(-1)^{k+1}}{k}=\sum_{k=m+1}^{2m}\frac{1}{k}+\frac{1}{2m+1}-\frac{1}{2m+2}\]
\[\sum_{k=m+1}^{2m}\frac{1}{k}+\frac{1}{2m+1}+\frac{1}{2m+2}-\frac{1}{m+1}\]
\[=\sum_{k=m+2}^{2(m+1)}\frac{1}{k}\]which is what you wanted

Answer 60

that is the proof, but maybe not the understanding of the steps

Answer 61

The only step that's troubling me now is the one where you use algebra to rewrite 1/(2m+1)-1/(2m+2) to the expression in the middle. :/ How do you realize that you're supposed to do that and how do you actually do it?

Answer 62

ok i will say it in english, bacause it will take me forever two write it

Answer 63

i realized it because it had to be true in order for this to work

Answer 64

Haha, sure.

Answer 65

in other words, i compared the sum \[\sum_{k=m+2}^{2(m+1)}\frac{1}{k}\]which is what you wanted to get, with the sum \[\sum_{k=m+1}^{2m}\frac{1}{k}\] which is what you had

Answer 66

i wrote what was in the first, that was not in the second

Answer 67

as i said, the first has the extra two terms \(\frac{1}{2m+1}\) and \(\frac{1}{2m+2}\) and was missing the first term \(\frac{1}{m+1}\)

Answer 68

then in order for this to work, it had to be true that
\[\frac{1}{2m+1}+\frac{1}{2m+2}-\frac{1}{m+1}=\frac{1}{2m+1}-\frac{1}{2m+2}\]

Answer 69

because by induction, you had
\[\sum_{k=m+1}^{2m}\frac{1}{k}+\frac{1}{2m+1}-\frac{1}{2m+2}\]

Answer 70

Could you reason like this: because the sum does not involve a negative end term because of 1/k that it "has" to be positive and for you to be able to "bake" it in you also have to take in account the term that the result that you're trying to get to "misses" precisely 1/(m+1)?

Answer 71

no

Answer 72

well, pellet. Haha

Answer 73

oh maybe
i did not think about negative numbers, but no matter because the algebra works

Answer 74

any way it was a matter of looking at what you needed to get what you wanted
if it was going to be true, it had to work

Answer 75

I'm just thinking about "how to think" in order for it to be viable to arrive at the correct conclusion, if you follow.

Answer 76

yes but inductive proofs are usually more mechanical than intuitive
they do not really give a reason, they just prove it is true

Answer 77

it is more or less obvious for example that
\[\sum_{k=1}^n=\frac{n(n+1)}{2}\] but the easy inductive proof does not explain how to get the formula at all, it just shows it is true for all \(n\)

Answer 78

gotta run, good luck

Answer 79

To arrive at a formula is more or less not supposed to be required, I'll just hope for the best in the future. Thanks so much for your time, and have a nice day/week/future. Adieu! =)