evaluate the limit of
[1/(n+1)+1/(n+2)+....+1/(n+n)
as n tends to infinity
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google is your friend. By searching for "sum 1/(n+k)" I found
http://math.stackexchange.com/questions/285308/limit-of-the-sum-frac1n1-frac1n2-cdots-frac12n
To evaluate the given limit, we can use the concept of Riemann sums and the integral.
First, let's rewrite the given expression using sigma notation:
∑ (1 / (n + k)), where k ranges from 1 to n
To proceed, we can convert the sum into an integral by considering it as a Riemann sum. We can rewrite the sum as follows:
∑ (1 / (n + k)) = 1/n * ∑ (1 / (1 + (k / n)))
As n approaches infinity, the term (k / n) approaches zero, so we can replace it with t, resulting in:
1/n * ∑ (1 / (1 + t))
We can convert the sum into an integral, replacing ∑ with ∫, and 1/n with Δt, where Δt represents the change in t:
∫ (1 / (1 + t)) dt
To integrate this expression, we can use the natural logarithm function:
∫ (1 / (1 + t)) dt = ln|1 + t| + C
Applying the limits of integration, we have:
ln|1 + t| |[0, n]
Let's substitute t back in terms of k:
ln|1 + (k / n)| |[0, n]
Now, let's evaluate the limit as n approaches infinity.
As n tends to infinity, the upper limit of integration (n) becomes large, so let's evaluate the upper limit:
ln|1 + (n / n)| = ln|2|
Similarly, let's evaluate the lower limit:
ln|1 + (0 / n)| = ln|1| = 0
Therefore, the limit of the given expression as n approaches infinity is:
ln|2|
Hence, lim[n→∞] ∑ (1 / (n + k)) = ln|2|.