If `t=(1^(2)+2^(2)+3^(2)+...r^(2))/(1^(3)+2^(3)+3^(3)+....+r^(3)), S_(n) = overset(n) underset(r=1) sum(-1)^(r)t_(r)` then `lim_(n to oo) ((1)/(3)-S_(n))=`

Explanation

To solve the problem, we need to find the limit \( \lim_{n \to \infty} \left( \frac{1}{3} - S_n \right) \), where \( S_n = \sum_{r=1}^{n} (-1)^r t_r \) and \( t_r = \frac{1^2 + 2^2 + 3^2 + \ldots + r^2}{1^3 + 2^3 + 3^3 + \ldots + r^3} \).

### Step 1: Calculate \( t_r \)

The formula for the sum of squares of the first \( r \) natural numbers is:
\[ 1^2 + 2^2 + 3^2 + \ldots + r^2 = \frac{r(r+1)(2r+1)}{6} \]
The formula for the sum of cubes of the first \( r \) natural numbers is:
\[ 1^3 + 2^3 + 3^3 + \ldots + r^3 = \left( \frac{r(r+1)}{2} \right)^2 \]
Thus, we can express \( t_r \) as:
\[ t_r = \frac{\frac{r(r+1)(2r+1)}{6}}{\left( \frac{r(r+1)}{2} \right)^2} \]

### Step 2: Simplify \( t_r \)

Now we simplify \( t_r \):
\[ t_r = \frac{r(r+1)(2r+1)}{6} \cdot \frac{4}{r^2(r+1)^2} \]
This simplifies to:
\[ t_r = \frac{4(2r+1)}{6r(r+1)} = \frac{2(2r+1)}{3r(r+1)} \]

### Step 3: Calculate \( S_n \)

Now we substitute \( t_r \) into \( S_n \):
\[ S_n = \sum_{r=1}^{n} (-1)^r t_r = \sum_{r=1}^{n} (-1)^r \frac{2(2r+1)}{3r(r+1)} \]
This can be rewritten as:
\[ S_n = \frac{2}{3} \sum_{r=1}^{n} (-1)^r \frac{2r+1}{r(r+1)} \]

### Step 4: Simplify the summation

The term \( \frac{2r+1}{r(r+1)} \) can be split:
\[ \frac{2r+1}{r(r+1)} = \frac{2}{r} - \frac{1}{r+1} \]
Thus:
\[ S_n = \frac{2}{3} \sum_{r=1}^{n} (-1)^r \left( \frac{2}{r} - \frac{1}{r+1} \right) \]

### Step 5: Evaluate the limit

As \( n \to \infty \), the series converges. The alternating series converges to a limit. We can analyze the behavior:
\[ \lim_{n \to \infty} S_n = \frac{2}{3} \left( \sum_{r=1}^{\infty} (-1)^r \frac{2}{r} - \sum_{r=1}^{\infty} (-1)^r \frac{1}{r+1} \right) \]
The first series converges to \( -\ln(2) \) and the second series converges to \( -\ln(2) + 1 \).

Thus:
\[ \lim_{n \to \infty} S_n = \frac{2}{3} \left( -\ln(2) - (-\ln(2) + 1) \right) = \frac{2}{3} \cdot 1 = \frac{2}{3} \]

### Step 6: Final limit calculation

Finally, we compute:
\[ \lim_{n \to \infty} \left( \frac{1}{3} - S_n \right) = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3} \]

### Final Answer
\[ \lim_{n \to \infty} \left( \frac{1}{3} - S_n \right) = -\frac{1}{3} \]