infinite series

[logistic_guy]

here is the question

Study the infinite series \(\displaystyle \sum_{n=1}^{\infty}\frac{n}{(n+1)!}\) and show that it converges to \(\displaystyle 1\).


my attemb
\(\displaystyle \sum_{n=1}^{\infty}\frac{n}{(n+1)!} = \frac{1}{(1 + 1)!} + \frac{2}{(2 + 1)!} + \frac{3}{(3 + 1)!} + \cdots = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots\)

\(\displaystyle = \frac{1}{2} + \frac{2}{6} + \frac{3}{24} + \cdots = \frac{1}{2} + \frac{1}{3} + \frac{1}{8} + \cdots\)

it don't look like it converge to \(\displaystyle 1\)

 

[Dr.Peterson]

Study the infinite series \(\displaystyle \sum_{n=1}^{\infty}\frac{n}{(n+1)!}\) and show that it converges to \(\displaystyle 1\).

my attempt
\(\displaystyle \sum_{n=1}^{\infty}\frac{n}{(n+1)!} = \frac{1}{(1 + 1)!} + \frac{2}{(2 + 1)!} + \frac{3}{(3 + 1)!} + \cdots = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots\)

\(\displaystyle = \frac{1}{2} + \frac{2}{6} + \frac{3}{24} + \cdots = \frac{1}{2} + \frac{1}{3} + \frac{1}{8} + \cdots\)

it don't look like it converge to \(\displaystyle 1\)

Why not? How did you "study" it?

Try writing out the first few partial sums. Then see if you can prove that what you see will continue.

 

[logistic_guy]

thank Dr.

Why not?

because i just don't see it

How did you "study" it?

by writing the first few terms

Try writing out the first few partial sums. Then see if you can prove that what you see will continue.

\(\displaystyle S_1 = \frac{1}{2}\)

\(\displaystyle S_2 = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}\)

\(\displaystyle S_3 = \frac{5}{6} + \frac{1}{8} = \frac{23}{24}\)

\(\displaystyle S_4 = \frac{23}{24} + \frac{1}{30} = \frac{119}{120}\)

am i have to suppose to see something here? i don't see anything
do i continue write more sums?

 

[Dr.Peterson]

am i have to suppose to see something here? i don't see anything
do i continue write more sums?

Either that, or just really look at those numbers you already have and think ...

That's what "study" means.

 

[logistic_guy]

or just really look at those numbers you already have and think ...

i think and i don't see anything

Either that,

let me try the next partial sum

\(\displaystyle S_5 = \frac{119}{120} + \frac{1}{144} = \frac{719}{720}\)

\(\displaystyle S_6 = \frac{719}{720} + \frac{1}{840} = \frac{5039}{5040}\)

\(\displaystyle S_7 = \frac{5039}{5040} + \frac{1}{5760} = \frac{40319}{40320}\)

\(\displaystyle S_8 = \frac{40319}{40320} + \frac{1}{45360} = \frac{362879}{362880}\)

now i realize something but i don't think it's strong evidence the sum converge to \(\displaystyle 1\)

i start with \(\displaystyle S_1 = \frac{1}{2} = 0.5\)

i end with \(\displaystyle S_8 = \frac{362879}{362880} = 0.999997\)

it seem it's getting close to one

do this mean i complete the show or i've to show more sums?

 

[Dr.Peterson]

Do you really not see the pattern?

Compare the numerators and denominators.

Of course, seeing the pattern doesn't prove the pattern, so you'll have some more work to do.

(The decimal values are evidence that it might converge to 1, without giving a clue about proving it.)

 

[logistic_guy]

Do you really not see the pattern?

Compare the numerators and denominators.

i see the difference between numator and denomator is one but i can't describe it in a formula

Of course, seeing the pattern doesn't prove the pattern, so you'll have some more work to do.

(The decimal values are evidence that it might converge to 1, without giving a clue about proving it.)

do might mean all my work of partial sum is useless

 

[Dr.Peterson]

do[es] might mean all my work of partial sum is useless

It means that listing decimal values gives you reason to think the claim might be true (which can encourage you to continue trying), but nothing more. Seeing the pattern in the exact fractional values gives you something specific to hypothesize, and then try to prove. Both have some value, but the latter has far more.

If you want to actually solve this problem, then try writing a formula for the apparent nth term (it isn't hard if you know even a tiny amount of algebra), and try proving it by mathematical induction. If not, I've said all I need to say.

 

[logistic_guy]

you want to actually solve this problem, then try writing a formula for the apparent nth term

i try but i can't write a formula

(it isn't hard if you know even a tiny amount of algebra)

not only tiny i'm very good in algebra

If not, I've said all I need to say.

i follow everything you say still can't write a formula
i think to see a a pattern and to invent formula base on that is a very high skill i don't have
my skills in algebra help me solve many difficult questions yet i'm lost in this one

if first term is \(\displaystyle \frac{1}{2}\) i can think of a formula \(\displaystyle \frac{n}{2n}\)
second and third terms will give me idea to correct my mistake and discover the \(\displaystyle n\)th term as \(\displaystyle \frac{n}{(n+1)!}\)
which is like going back to square \(\displaystyle 1\)

for the sum it's very diffcult to write the \(\displaystyle n\)th sum
i can't see a relation between \(\displaystyle \frac{1}{2}\) and \(\displaystyle \frac{5}{6}\) which also between \(\displaystyle \frac{5}{6}\) and \(\displaystyle \frac{23}{24}\)

the difference between numbers don't help
the division between numbers don't help
the square root between numbers don't help
the only thing i can see the denomator can work with factorial but that don't help because the original summation is also factorial
so the formula for the \(\displaystyle n\)th partial some is something like \(\displaystyle \frac{*}{(1 + n)!}\)
the star* represent the relationship between all numators which is unknow for now