Telescoping : Product it or sum it

Quite simply (without algebra), the first term of the product is $\frac12$ and then you are multiplying by numbers which are all less than 1. Therefore the product can never be greater than $\frac12$ so does not equal 1

$\begin{aligned} P & = \frac 12 \times \frac 23 \times \frac 34 \times \cdots \\ & = \lim_{n \to \infty} \frac 12 \times \frac 23 \times \frac 34 \times \cdots \times \frac {n-2}{n-1} \times \frac {n-1}n \times \frac n{n+1} \\ & = \lim_{n \to \infty} \frac 1{\color{#3D99F6} \cancel 2} \times \frac {\color{#3D99F6}\cancel 2}{\color{#D61F06}\cancel 3} \times \frac {\color{#D61F06}\cancel 3}{\color{#3D99F6}\cancel 4} \times \cdots \times \frac {\color{#3D99F6}\cancel {n-2}}{\color{#D61F06}\cancel {n-1}} \times \frac {\color{#D61F06}\cancel {n-1}}{\color{#3D99F6}\cancel n} \times \frac {\color{#3D99F6}\cancel n}{n+1} \\ & = \lim_{n \to \infty} \frac 1{n+1} \\ & = 0 \ne 1 \end{aligned}$

$\implies$ No , the limit is not 1 but 0.

Before we deal with the infinite telescoping series , $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ we must be careful about the partial sum of the series. The partial sum of the telescoping series above can be written as $\sum_{n=1}^{m} \frac{1}{n(n+1)} = 1 - \frac{1}{m+1}$ Now when $m\to \infty$ then $\lim_{m\to\infty} \sum_{n=1}^{m} \frac{1}{n(n+1)} =1 - \frac{1}{m+1} = 1 - \frac{1}{\infty +1} = 1-0 = 1$

On the same way the partial product of telescoping product when $m\to \infty$ is $\lim_{m\to\infty} \prod_{n=1}^{m} = 1\times \frac{1}{m+1} = 1 \times \frac{1}{\infty \times m } = 0$

Hence, the answer is No .

NO.

Let's say that both series ends with infinity which is x.

The telescoping sum would become

1 = 1 - 1/2 + 1/2 - ... + 1/(x-1) - 1/x + 1/x - 1/(x+1)

Everything would cancel and we will be left with

1 = 1 - 1/(x+1)

A number added to infinity is still infinity,

1 = 1 - 1/x

1 divided by infinity would grow smaller and smaller until it can just become 0.

1 = 1 - 0

Proving the telescoping sum correct.

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But, for the telescoping product, we have

1 = (1/2)(2/3)(3/4)...((x-1)/x)(x/(x+1))

Everything would cancel until we have

1 = 1/(x+1)

As we said before, infinity added to anything is infinity, and anything divided by infinity is 0.

1 = 0

The answer to the telescoping product is 0, not 1.

Hence, the answer is no.

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You could also look at it by multiplying both sides by (x+1), and we will get

x+1 = 1

Since x is not 0, this equation will not work.

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(my cheap version)

Please correct me if I’m wrong here, but I saw that the above could be rewritten as $\frac {n!}{(n+1)!}$ ; thus, it is clear that $lim_{n\to\infty} \frac {n!} {(n+1)!} = 0$ . Is this oversimplified? Do let me know!

If you shift just the denominators 1 place to the right (and loop n around to the first place) you get:

$\frac{1}{n} \times \frac{2}{2} \times \frac{3}{3} \times \frac{4}{4} \times ...$

The product of the first two terms is 1/3, and the product of that and the third term is 1/4, and the product of that and the fourth term is 1/5, and so on, for a product that is the limit of 1/n as n approaches infinity: Zero.

We get the infinite sum as 0 which is not equal to 1

Each subsequent term after 1/2 makes the result smaller. So the series cannot tend to 1.

The denumerator of the last term doesn't get cancelled.

If you multiply 1/2 by any number less than one (which would include all proper fractions), the product will be less than 1/2. Multiplying by additional proper fractions reduces the value further. Therefore, since the product will always be less than 1/2, then the product can never be equal to 1.

Less than 1 × less than 1 will always be less than itself, such as .9×.9= .81, which is less than .9. An infinite amount of "times" would reach 0 (or infinitely close to 0).

I don’t think you can get 1 by times-ing by numbers less than 1

Even if all the fractions in the middle cancel out, there will be the 1 at the beginning and an infinitely large number as a factor in the denominator that does not get cancelled. Therefore the answer should be 1 divided by an infinitely large number equals 0.

The expression is in the form :

$\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} .......... \times \frac{n - 1}{n}$

So it is clear that every term get's cancelled except the first term that is 1 and the last term that is n .

Therefore the expression doesn't converge to 1 .

All of the numbers in the second one are fractions less than $1$ . The product of any numbers that are all less than one cannot equal one.

It can never be one as the product of the two numerators will always be less than the product of the denominators.

Somewhere (inf.-1)÷ inf. will come so it will become (inf.-1)×(1÷inf.), so the product will become 0.

In such a product the number we divide by is constantly growing. Thus, finally (if there can be a "final") we get one, divided by infinity. Such a number is very close to zero, and it cannot converge to 1 anyway. Therefore the right answer is "no". P.S. Sorry for my English if I made a mistake somewhere. Russian is my native language.

For the sum, it would be 1. But for the product, it would be 1/2 2/3 3/4 ... [(n-1)/n]*[n/(n+1)]=1/(n+1) which is definitely not equal to 1.

The maximum value is (n-1)/n which practically results in 0.99999999999.. .. which is not equal '1'

I hope this is right, but if you use logarithms you get: log $\frac{1}{2}$ log $\frac{2}{3}$ log $\frac{3}{4}$ whic equals: log 1(=0) - log 2 + log 2 - log 3 + log 3... where you can cancel all the logarithms except log 1 which is zero

1/2 x 2/3 becomes two thirds of a half. That multiplied by 3/4 becomes three quarters of two thirds of a half. That pattern goes on. In the end, the final result will be close to a half, but not approach half. With each multiplication step, the number goes further from 1/2, but each step away from 1/2 is smaller. Hence, the result is a bit close to 1/2, but it never nears it.