Evaluate the following
(i) `lim_(n to oo)((1)/(n^(2))+(2)/(n^(2))+(3)/(n^(2))....+(n-1)/(n^(2)))`
(ii) `lim_(n to oo)((1)/(n+1)+(1)/(n+2)+....+(1)/(2n))`
(iii) `lim_(n to oo)((n)/(n^(2)+1^(2))+(n)/(n^(2)+2^(2))+....+(n)/(2n^(2)))`
(iv) `lim_(n to oo)((1^(p)+2^(p)+.....+n^(p)))/(n^(p+1)),pgt0`

Answer

To evaluate the given limits, we will follow a systematic approach for each part.

### (i) Evaluate
\[ \lim_{n \to \infty} \left( \frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2} + \ldots + \frac{n-1}{n^2} \right) \]

**Step 1: Rewrite the sum**
We can express the sum as:
\[ \sum_{r=1}^{n-1} \frac{r}{n^2} \]

**Step 2: Factor out \(\frac{1}{n^2}\)**
This gives us:
\[ \frac{1}{n^2} \sum_{r=1}^{n-1} r \]

**Step 3: Use the formula for the sum of the first \(n-1\) integers**
The formula for the sum of the first \(k\) integers is \(\frac{k(k+1)}{2}\). Thus,
\[ \sum_{r=1}^{n-1} r = \frac{(n-1)n}{2} \]

**Step 4: Substitute back into the limit**
Now substituting this back, we have:
\[ \frac{1}{n^2} \cdot \frac{(n-1)n}{2} = \frac{(n-1)}{2n} \]

**Step 5: Take the limit as \(n\) approaches infinity**
\[ \lim_{n \to \infty} \frac{(n-1)}{2n} = \lim_{n \to \infty} \frac{1 - \frac{1}{n}}{2} = \frac{1}{2} \]

### Final Answer for (i):
\[ \frac{1}{2} \]

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### (ii) Evaluate
\[ \lim_{n \to \infty} \left( \frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} \right) \]

**Step 1: Rewrite the sum**
This can be expressed as:
\[ \sum_{r=1}^{n} \frac{1}{n+r} \]

**Step 2: Factor out \(\frac{1}{n}\)**
This gives us:
\[ \frac{1}{n} \sum_{r=1}^{n} \frac{1}{1 + \frac{r}{n}} \]

**Step 3: Recognize it as a Riemann sum**
As \(n\) approaches infinity, this sum approaches the integral:
\[ \int_0^1 \frac{1}{1+x} \, dx \]

**Step 4: Evaluate the integral**
The integral can be computed as:
\[ \int_0^1 \frac{1}{1+x} \, dx = \ln(1+x) \bigg|_0^1 = \ln(2) - \ln(1) = \ln(2) \]

### Final Answer for (ii):
\[ \ln(2) \]

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### (iii) Evaluate
\[ \lim_{n \to \infty} \left( \frac{n}{n^2 + 1^2} + \frac{n}{n^2 + 2^2} + \ldots + \frac{n}{n^2 + (2n)^2} \right) \]

**Step 1: Rewrite the sum**
This can be expressed as:
\[ \sum_{r=1}^{2n} \frac{n}{n^2 + r^2} \]

**Step 2: Factor out \(\frac{1}{n}\)**
This gives us:
\[ \sum_{r=1}^{2n} \frac{1}{1 + \left(\frac{r}{n}\right)^2} \]

**Step 3: Recognize it as a Riemann sum**
As \(n\) approaches infinity, this sum approaches the integral:
\[ \int_0^2 \frac{1}{1+x^2} \, dx \]

**Step 4: Evaluate the integral**
The integral can be computed as:
\[ \int_0^2 \frac{1}{1+x^2} \, dx = \tan^{-1}(x) \bigg|_0^2 = \tan^{-1}(2) - \tan^{-1}(0) = \tan^{-1}(2) \]

### Final Answer for (iii):
\[ \tan^{-1}(2) \]

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### (iv) Evaluate
\[ \lim_{n \to \infty} \frac{1^p + 2^p + \ldots + n^p}{n^{p+1}}, \quad p > 0 \]

**Step 1: Rewrite the sum**
This can be expressed as:
\[ \frac{1}{n^{p+1}} \sum_{r=1}^{n} r^p \]

**Step 2: Use the formula for the sum of powers**
The formula for the sum of the first \(n\) powers is approximately \(\frac{n^{p+1}}{p+1}\) for large \(n\). Thus,
\[ \sum_{r=1}^{n} r^p \sim \frac{n^{p+1}}{p+1} \]

**Step 3: Substitute back into the limit**
Now substituting this back, we have:
\[ \frac{1}{n^{p+1}} \cdot \frac{n^{p+1}}{p+1} = \frac{1}{p+1} \]

### Final Answer for (iv):
\[ \frac{1}{p+1} \]

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