Evaluate the limit of the sequence

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Homework Statement


2vx2eqw.jpg
[/B]

Homework Equations


delta(x) = [b-a]/n
xi=a+delta(x)i

The Attempt at a Solution


So, it is said that i have to use riemann sums to solve this one.
what i did is i took the 1/12k out thus getting

1/12k[1/[1+1/12k] + 1/[1+2/12k] + 1/{1+19k/12k]]

I found that

xi = 1+1/12k
so ,
b-a = 1/12 and since a=1 b=13/12

i find that f(x) = 1/x and i evalute the integral from 1 to 12/12 of 1/x

I keep getting the wrong answer i really don't know what to do anymore
 
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gkamal said:

Homework Statement


2vx2eqw.jpg
[/B]

Homework Equations


delta(x) = [b-a]/n
xi=a+delta(x)i

The Attempt at a Solution


So, it is said that i have to use riemann sums to solve this one.
what i did is i took the 1/12k out thus getting

1/12k[1/[1+1/12k] + 1/[1+2/12k] + 1/{1+19k/12k]]
This isn't right. Above you show three terms.
b1 is a sum that contains 8 terms. How many terms are there in b2? b3? bn?
gkamal said:
I found that

xi = 1+1/12k
so ,
b-a = 1/12 and since a=1 b=13/12

i find that f(x) = 1/x and i evalute the integral from 1 to 12/12 of 1/x
Since 12/12 = 1, the integral is ##\int_1^1 \frac 1 x dx = 0##.
gkamal said:
I keep getting the wrong answer i really don't know what to do anymore
 
Mark44 said:
This isn't right. Above you show three terms.
b1 is a sum that contains 8 terms. How many terms are there in b2? b3? bn?
Since 12/12 = 1, the integral is ##\int_1^1 \frac 1 x dx = 0##.
i found the answer already but still thanks btw the 12/12 was a typo obviously since the b found on the line before was 13/12
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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