`lim_(n->oo)(1/(n^2+1)+2/(n^2+2)+3/(n^2+3)+....n/(n^2+n))`

Answer

To solve the limit \( L = \lim_{n \to \infty} \left( \frac{1}{n^2 + 1} + \frac{2}{n^2 + 2} + \frac{3}{n^2 + 3} + \ldots + \frac{n}{n^2 + n} \right) \), we can follow these steps:

### Step 1: Rewrite the expression
We can express the limit as:
\[ L = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{n^2 + k} \]

### Step 2: Analyze the general term
For large \( n \), the term \( n^2 + k \) can be approximated by \( n^2 \) since \( k \) is much smaller than \( n^2 \). Thus, we can rewrite the term:
\[ \frac{k}{n^2 + k} \approx \frac{k}{n^2} \]

### Step 3: Substitute the approximation into the sum
Now, substituting this approximation into our sum gives:
\[ L \approx \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{n^2} = \lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^{n} k \]

### Step 4: Calculate the sum of the first \( n \) natural numbers
The sum of the first \( n \) natural numbers is given by:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
Substituting this into our expression gives:
\[ L \approx \lim_{n \to \infty} \frac{1}{n^2} \cdot \frac{n(n + 1)}{2} = \lim_{n \to \infty} \frac{n^2 + n}{2n^2} \]

### Step 5: Simplify the expression
Now we can simplify:
\[ L = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2} = \frac{1 + 0}{2} = \frac{1}{2} \]

### Conclusion
Thus, the limit is:
\[ \boxed{\frac{1}{2}} \]