Finding value of a series in R without for-loop

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I am a newbie in R` and I found this problem:

Calculate the following sum using R: 1+(2/3)+(2/3)(4/5)+...+(2/3)(4/5)...(38/39)

I was enthusiastic to know how to solve this without using a for loop, and using only vector operations.

My thoughts and what I've tried till now:

Suppose I create two vectors such as

x<-2*(1:19)
y<-2*(1:19)+1

Then, x consists of all the numerators in the question and y has all the denominators. Now

z<-x/y

will create a vector of length 19 in which will be stored the values of 2/3, 4/5, ..., 38/39

I was thinking of using the prod function in R to find the required products. So, I created a vector such that

i<-1:19

In hopes of traversing z from the first element to the last, I did write:

prod(z[1:i])

But it failed miserably, giving me the result:

[1] 0.6666667
Warning message:
In 1:i : numerical expression has 19 elements: only the first used

What I wanted to do:

I expected to store the values of (2/3), (2/3)(4/5), ..., (2/3)(4/5)...(38/39) individually in another vector (say p) which will thus have 19 elements in it. I then intend to use the sum function to finally find out the sum of all those...

Where am I stuck:

As described in the R documentation, the prod function returns the product of all the values present in its arguments. So,

prod(z[1:1])
prod(z[1:2])
prod(z[1:3])

will return the values of (2/3), (2/3)(4/5), (2/3)(4/5)(6/7) respectively which it does:

> prod(z[1:1])
[1] 0.6666667
> prod(z[1:2])
[1] 0.5333333
> prod(z[1:3])
[1] 0.4571429

But it's not possible to go on like this and do it for all the 19 elements of the vector z. I am stuck here thinking as to what could be done. I wanted to iterate all the elements of z one-by-one for which I created another vector i as described above, but it didn't go as I had thought. Any help, suggestions, and hints will be really great as to how this can be done. I seem to have run out of ideas here.

More Information:

Here, I am providing with all the outputs in a systematic manner for others to understand my problem better:

> x
 [1]  2  4  6  8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38
> y
 [1]  3  5  7  9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39
> z
 [1] 0.6666667 0.8000000 0.8571429 0.8888889 0.9090909 0.9230769 0.9333333
 [8] 0.9411765 0.9473684 0.9523810 0.9565217 0.9600000 0.9629630 0.9655172
[15] 0.9677419 0.9696970 0.9714286 0.9729730 0.9743590
> i
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19

Short Note (controversial statement ahead): This post would really have benefited from the use of LaTeX, but unfortunately, due to extremely heavy dependencies, as is mentioned in several posts regarding inclusion of LaTeX on Stack Overflow (like this), that is not a thing till now.

vector r
1 Answers

You can use cumprod to get a cumulative product of a vector which is what you are after

p <- cumprod(z)
p
# [1] 0.6666667 0.5333333 0.4571429 0.4063492 0.3694084 0.3409923 0.3182595
# [8] 0.2995384 0.2837732 0.2702602 0.2585097 0.2481694 0.2389779 0.2307373
# [15] 0.2232941 0.2165276 0.2103411 0.2046562 0.1994087

A less-efficient but more generalized alternative to cumprod would be

p <- sapply(i, function(x) prod(z[1:x]))

Here the sapply takes the place of the loop and passes a different ending index for each product

Then you can do

1 + sum(p)
user_23720640
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