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 Sum Exponent Ratio Evaluation (Posted on 2023-05-27)
Evaluate:
```        n + n2 + n3 +...  ...+ nn
Limit  --------------------------
nāā    1n + 2n + 3n +...  ...+ nn```

 Submitted by K Sengupta Rating: 5.0000 (1 votes) Solution: (Hide) ```(Expression on numerator)k = n+n^2+n^3+........+n^n) = n(1+n+n^2+....+n^(n-1)) n^n-1 = n*--x------ n-1 Therefore, (expression on the numerator)/n^n n 1 = -----* (1- ---- )k n-1 n^n n 1 Or, lim ----- * (1- ---- ) n-> infinity n-1 n^n = 1*1 = 1 Lim (Expression on the denominator)/n^n n-> infinity 1 2 (n-1)^n n^n = lim ----- + ----+..........+ --------+ ------- n -> ∞ n^n n^n n^n n^n = lim (1^n + (1-1/n)^n +(1-2/n)^n+.......) n-> infinity Now, lim (1^n + (1-1/n)^n +(1-2/n)^n+.......) n-> infinity = 1+ 1/e + 1/e^2+....... 1 e =----- = ------- 1-1/e e-1 Consequently, required limit = 1/(e/e-1) = (e-1)/e = 1- 1/e```

 Subject Author Date Got it with some help Larry 2023-05-27 14:35:27 numeric approximation Larry 2023-05-27 10:50:01

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