Consider first the sum alone, without taking a limit.

```
var('k,n')
E = (3*n)/sqrt(2*n^4 - k^2*n^2)
S = sum(E.canonicalize_radical(), k, 1, n)
```

The value of `S`

is more or less an echo.
In other words, sage has no idea (and no chance) to find a closed formula for the sum.

```
sage: S
3*sum(1/sqrt(-k^2 + 2*n^2), k, 1, n)
```

Taking the limit from such an expression produces an other echo.

```
sage: S.limit(n = oo)
3*limit(sum(1/sqrt(-k^2 + 2*n^2), k, 1, n), n, +Infinity)
```

In such cases, we can ask for numerical values for a "big $n$":

```
sage: n = 10**6; 3.0 * sum([ 1.0 / sqrt(2.*n^2 - k^2) for k in range(1, n + 1)])
2.35619492953245
```

For such numerical computations pari/gp does a better job, we have for instance

```
? \p 100
realprecision = 115 significant digits (100 digits displayed)
? n = 10^6; sum(k=1, n, 3 / sqrt(2*n^2 - k^2))
%2 = 2.356194929532423149025695836193084006094556217514368071115402460936426617919755947726804564160237427
? n = 10^8; sum(k=1, n, 3 / sqrt(2*n^2 - k^2))
%3 = 2.35619449458574323604876967144696073157514146025901053849933599052885950787940136378309129299648754
```

Mathematically we have a masked Riemann sum:
$$
3\lim_{n\to\infty}\sum_{1\le k\le n}\frac 1{\sqrt{2n^2-k^2}}
= 3\lim_{n\to\infty}\frac 1n\sum_{1\le k\le n}\frac 1{\sqrt{2-(k/n)^2}}
= 3\int_0^1\frac{dx}{2-x^2}
= \frac{3\pi}4\ ,
$$
and this involved integral can be easily done in sage:

```
sage: var('x');
sage: integral( 3 / sqrt(2 - x^2), x, 0, 1)
3/4*pi
sage: _.n()
2.35619449019234
```

Sage is stronger, when the typist acts both as a good mathematician and as a good programmer, and it was designed to work best in such a combination.