Ta có :

\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\)

\(=1-\frac{1}{2013}< 1\)( đpcm )

\(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

....

\(\frac{1}{2013^2}< \frac{1}{2012.2013}=\frac{1}{2012}-\frac{1}{2013}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}=1-\frac{1}{2013}< 1\)

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}\)

Ta thấy \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2013^2}\)\(< \frac{1}{2012.2013}\)

\(\Rightarrow A< B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\)

MÀ \(B=1-\) \(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...-\frac{1}{2012}+\frac{1}{2012}-\frac{1}{2013}\)

        \(=1-\frac{1}{2013}< 1\)

Mà A<B nên A<1(t/c)(đpcm)

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2013^2}\)

Ta có ; 

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

...

\(\dfrac{1}{2013^2}< \dfrac{1}{2012.2013}\)

\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2012.2013}\)

\(\Rightarrow B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2013}\)

\(\Leftrightarrow B< 1-\dfrac{1}{2013}\)

\(\Rightarrow B< \dfrac{2012}{2013}\)

Lại có : \(\dfrac{2012}{2013}< \dfrac{3}{4}\)

\(\Rightarrow B< \dfrac{3}{4}\)

* Chắc vậy, sai thì thôg cảm ^^ * 

Còn j k hiểu thì ib nha

Xét số hạng tổng quát: \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\) (do \(\sqrt{n+1}-\sqrt{n}>0\forall n\in\mathbb{N}\text{ nên ta có thể nhân liên hợp}\))

Áp dụng vào và ta có:

\(VT=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2013^2}-\sqrt{2013^2-1}\)

\(=\sqrt{2013^2}-1=2013-1=2012^{\left(đpcm\right)}\)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{2013^2}\\ =\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{2013.2013}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}\\ -1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2012}-\dfrac{1}{2013}\\ =1-\dfrac{1}{2013}< 1\)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}=1-\dfrac{1}{2}+\dfrac{1}{2}+....+\dfrac{1}{2012}-\dfrac{1}{2013}\)

\(\Rightarrow dpcm\)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2012}-\dfrac{1}{2013}=1-\dfrac{1}{2013}< 1\left(đpcm\right)\)

ủa, nó nhỏ hơn 1 mà sao bn lại ghi lớn hơn 1

\(\frac{1}{1.2}+\frac{2}{1.2.3}+\frac{3}{1.2.3.4}+...+\frac{2013}{1.2...2014}\)

\(=\frac{1}{2}+\frac{1}{1.3}+\frac{1}{1.2.4}+...+\frac{1}{1.2...2012.2014}\)

\(=\frac{1.1.3.4...2012.2014}{2.1.3.4...2012.2014}+\frac{1.2.4.5...2012.2014}{1.3.2.4.5...2012.2014}+...+\frac{1}{1.2.....2012.2014}\)(Quy đồng mẫu)

\(=\frac{1.1.3.4...2012.2014+1.2.4.5...2012.2014+...+1}{1.2...2012.2014}>1\)

ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều 

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