Evaluate the following limits :
`Lim_(n to oo) (Sigman^(3))/(n^(4))`

Answer

To evaluate the limit

\[ \lim_{n \to \infty} \frac{\sum_{k=1}^{n} k^3}{n^4}, \]

we will follow these steps:

### Step 1: Find the formula for the summation of cubes
The formula for the sum of the first \( n \) cubes is given by:

\[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2. \]

### Step 2: Substitute the summation into the limit
Substituting the formula into our limit, we have:

\[ \lim_{n \to \infty} \frac{\left(\frac{n(n+1)}{2}\right)^2}{n^4}. \]

### Step 3: Simplify the expression
We can simplify the expression inside the limit:

\[ = \lim_{n \to \infty} \frac{n^2(n+1)^2}{4n^4}. \]

This simplifies to:

\[ = \lim_{n \to \infty} \frac{(n^2 + 2n + 1)}{4n^2}. \]

### Step 4: Divide each term by \( n^2 \)
Now, we can divide each term in the numerator by \( n^2 \):

\[ = \lim_{n \to \infty} \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{4}. \]

### Step 5: Evaluate the limit as \( n \to \infty \)
As \( n \) approaches infinity, \( \frac{2}{n} \) and \( \frac{1}{n^2} \) both approach 0. Therefore, we have:

\[ = \frac{1 + 0 + 0}{4} = \frac{1}{4}. \]

### Final Answer
Thus, the limit evaluates to:

\[ \lim_{n \to \infty} \frac{\sum_{k=1}^{n} k^3}{n^4} = \frac{1}{4}. \]

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