Solving Limits with a Clear Pattern: Techniques and Strategies for Success

[Dell]

how would you go about solving these :

lim
n->inf (1/2 + 3/2² + 5/2³ +... +(2n-1)/2ⁿ)

and

lim
n->inf (1²/n³+3²/n³+5²/n³+...+(2n-1)²/n³)

the pattern is very clear but what is the technique i should use to solve these?

 

[tiny-tim]

Hi Dell!

(have an infinity: ∞ )

lim
n->inf (1/2 + 3/2² + 5/2³ +... +(2n-1)/2ⁿ)

Try fiddling about with it until you can integrate it to something you can sum easily

 

[Mark44]

lim
n->inf (1²/n³+3²/n³+5²/n³+...+(2n-1)²/n³)

For this one you can factor n³ out of all of the denominators (but not out of the summation). Then you're left with finding a closed form representation of 1² + 3² + 5² + ... + (2n - 1)². This would be equal to the sum of the squares of the positive integers up to (2n), with the sum of the squares of the even integers up to (2n) subtracted off.

 

[Dell]

For this one you can factor n³ out of all of the denominators (but not out of the summation). Then you're left with finding a closed form representation of 1² + 3² + 5² + ... + (2n - 1)². This would be equal to the sum of the squares of the positive integers up to (2n), with the sum of the squares of the even integers up to (2n) subtracted off.

how do i find the sum of squares of integers

 

[Dell]

Hi Dell!

(have an infinity: ∞ )


Try fiddling about with it until you can integrate it to something you can sum easily

i see that the numerator is a series with a difference of +2
and the denominator is a series of *2

but i don't see how to simplify it

 

[tiny-tim]

Hint: what does (2n - 1)/(√2)²ⁿ remind you of?

 

[Dell]

do you mean that i should make t=2n and then integrate in parts? how would that help me? how would integration help at all?

this is what i have don't so far, but its wrong somewhere:


the sum S is

S = 1/2 + 3/4 + 5/8 + 7/16 +9/32 +... (2n-1)/2^n

now i take double that and i get
2S = 1 + 3/2 + 5/4 + 7/8 + 9/16 +...(4n-2)/2^n

now if i subtract 2S-S i get
2S = 1 + 3/2 + 5/4 + 7/8 + 9/16 +...(4n-2)/2^n
S =//// 1/2 + 3/4 + 5/8 + 7/16 +9/32 +... (2n-1)/2^n

as you see, if i subtract S from 2S all the middle fractions have a matching one (eg, 3/2 -1/2 =2/2 5/4 -3/4=2/4 etc) only the 1 from 2S and the (2n-1)/2^n from S are left with the sequence (2/2 +2/4 +2/8...)

2S – S = 1 + 2/2 + 2/4 + 2/8 + …2/2^n - (2n-1)/2^n

S = 1 + 2*( 1/2 + 1/4 + 1/8 …) - (2n-1)/2^n

now i know that 2*( 1/2 + 1/4 + 1/8 …) =2*1=2
so

S=1+2-(2n-1)/2^n
=3-(2n-1)/2^n


only that the answer is wrong, and the correct one is

3-(2n+3)/2^n

can you see where i have gone wrong??

 

[tiny-tim]

… as you see, if i subtract S from 2S all the middle fractions have a matching one (eg, 3/2 -1/2 =2/2 5/4 -3/4=2/4 etc) only the 1 from 2S and the (2n-1)/2^n from S are left with the sequence (2/2 +2/4 +2/8...)

ooh, that's clever … i didn't see that!

now i know that 2*( 1/2 + 1/4 + 1/8 …) =2*1=2
can you see where i have gone wrong??

ah … that line is correct if you're going to ∞,

but for some reason the question wants you only to go to the nth term, so it's = 2 - … ?

(btw, I was thinking of f(√2) with f(x) = ∑(2n-1)/x²ⁿ, so that ∫f(x) = -∑1/x^2n-1 … but your method is better )