Question

Please Help I'm stuck really bad
can someone help me show this equation = m please:
lim (((n+1)^m )-n^m)
n->+inf ---------------- = m
n^(m-1)

Answer 1

Perhaps this helps: (1 + a)^m = 1 + m a for a<<1

Answer 2

Honestly I'm just really confused by this problem, Ive tried looking at it different ways but i haven't been able to figure it out

Answer 3

there is a binomial result from (a+b)^n

\[\sum_{k=0}^{n}\binom{n}{k}a^{n}b^{n-k}\]

Answer 4

since b=1 in this case .... that reduces it visually as

\[\sum_{k=0}^{n}\binom{n}{k}a^{n}\]using n on my part was not a good idea since there is an n in your setup originally, just habit

Answer 5

the binomial (n k) = n!/[(n-k)!k!] = n(n-1)(n-2)...(n-k)...(3)(2)(1)
------------------------
(n-k)...(3)(2)(1) * k!


n(n-1)(n-2)...(n-k+1)
-------------------
k!

Answer 6

Alright so far I get it .. But Im just confused to what tipped you off to start this problem with the binomial law ?

Answer 7

that (n+1)^m as m to infinity, it is a binomial

Answer 8

ugh, its n to infinity lol

Answer 9

Lol cool, but then the rest is still confusing though .. Are you breaking them down in parts ?

Answer 10

since m is constant here, it makes the setup a little more easy to work with


n^m + n^(m-1) m + n^(m-2) m(m-1)(m-2)/2! + ... - n^m
---------------------------------------------------
n^(m-1)

Answer 11

n^(m-1) m + n^(m-2) m(m-1)(m-2)/2! + ...
-----------------------------------------
n^(m-1)


m + n^(m-2-m+1) m(m-1)(m-2)/2! + n^(m-3-m+1) m(m-1)(m-2)(m-2)/3!+...


m + n^(-1) m(m-1)(m-2)/2! + n^(-2) m(m-1)(m-2)(m-2)/3!+ n^(-3)...

all the n parts are now fractions, and as n goes to infinty they simply zero down, and we are left with that constant first term

Answer 12

\[\Large\frac{\left[\sum_{k=0}^{m}\binom{m}{k}n^k\right]-n^m}{n^{m-1}}\]
\[\Large\frac{\sum_{k=0}^{m-1}\binom{m}{k}n^k}{n^{m-1}}\]
\[\Large\sum_{k=0}^{m-1}\frac{\binom{m}{k}n^k}{n^{m-1}}\]
\[\Large\sum_{k=0}^{m-1} \binom{m}{k}n^{k-m+1}\]
looks about right to me ....

Answer 13

So you literally separated the binomial part .. solved it then used it to cancel out the rest of the terms ?

Answer 14

yes

Answer 15

the summation notation might be seen better if n is left on the bottom:

\[\Large\sum_{k=0}^{m-1} \binom{m}{k}\frac{1}{n^{m-1-k}}\]
as n to infinity, all the terms zero out except for that very last term: \[\binom{m}{m-1}\frac{1}{\infty^0}:=m\]

Answer 16

Alright I get it Thanks!! I just have a follow up question though .. Could we have solved the equation if we had done root of m ? I know it will cancel out all m's in the equation but I,m guessing that that's already a bad first step right ?

Answer 17

i am always leary of altering the setup in such an abrupt manner, it just tends to increase the chance for error to me.

Answer 18

spose we mth root it so that m is an even number, but the innards reduce to a negative; we have moved from the reals into the complex

Answer 19

Yeah I agree ... Therefore it's lot simpler using the sum identities

Answer 20

Thanks A lot!! You really saved me :D

Answer 21

it was a fun problem to delve into :)