N- N 1-1-2-1-3-1-n-Snthen Sn-

The correct option is C ≥√(n) s_n=n2[a+l] s_n=n2[1+1√(n)] s_n=n2[1+√(n)√(n)] s_n=n+√(n)2 s_n ≥√(n) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Sum of N Terms of an AP

n∈ N 1+1√2+1...

Question

$n \in N$
$1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . . . . . . + \frac{1}{\sqrt{n}} = S n$
then $S n = ?$

\sqrt{n}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

< \sqrt{n}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\leq \sqrt{n}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\geq \sqrt{n}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

Suggest Corrections

Similar questions

Statement I

n \geq 2

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} > \sqrt{n}

Statement II

n \geq 2

\sqrt{n (n + 1)} < n + 1

t_{n}

n^{t h}

S_{n}

n

t_{m} = \frac{1}{n}

t_{n} = \frac{1}{m}

S_{m n}

n \in N, 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} = S_{n}

S_{n}

S_{n} = 1 + 2 + 3 + . . . + n

P_{n} = \frac{S_{2}}{S_{2} - 1} \cdot \frac{S_{3}}{S_{3} - 1} \cdot \frac{S_{4}}{S_{4} - 1} \cdot \cdot \cdot \frac{S_{n}}{S_{n} - 1}

n \in N, (n \geq 2) .

lim n \to \infty P_{n} =

S_{n} = \frac{n}{(n + 1) (n + 2)} + \frac{n}{(n + 2) (n + 4)} + \frac{n}{(n + 3) (n + 6)} + . . . . . . + \frac{1}{6 n}

lim n \to \infty S_{n}

Join BYJU'S Learning Program